Basics of Physics
Measurements and dealing with their uncertainties are some first skills you learn in a lab. In this article, I will go over what you need to know when dealing with data.
Table of Contents
Measurement and Uncertainty
The beginning of physics started with measuring quantities. It soon became apparent that not all data meant the same. It was important to have reliable measurements. However, that was often a problem. Measuring things was often a random affair. The devices used were not all made the same either. Many measuring devices were poorly made and inaccurate at first. So measurements will never be perfect. They will not because humans are error prone and the devices we use are as well.
Uncertainty
Every device, like a ruler, has a built in amount of uncertainty. It is vital to state this when publishing measurements. It also gives the uncertainty as a percentage. This can help make it clear. It is the ratio of the uncertainty to the measured value, multiplied by 100.
If we measure a piece of metal to be 6.5 cm long, we can assume it has an uncertainty of .1 cm. To get the percentage of uncertainty we divide:
\[ \frac{0.1}{6.5} = .015 = 1% \]
Significant Figures
This is the number of reliably known digits in a number. Nonzero digits are usually significant. Zeroes vary according to the situation. In basic lab work, you never want to state more precision than you have. This is just inaccuracy. If your measurements have 3 significant digits, then your calculations should have only 3 significant digits. Many people do this, don’t be one of them, please. If you are dealing with numbers that contain different amounts of significant digits, then always use the least amount. Consider the number 12 and 23.4. I use these two numbers in a calculation. The result should have 2 significant figures in it.
Scientific Notation
When you are dealing with numbers that contain many digits it is convenient to use scientific notation. A number like 100,00,000 can be written as \( 1.0 * 10^7 \). This is just a standard way to display numerical values. Another example is the number .00000383. I would write it as \( 3.83 * 10^6 \).
Derived Quantities
Derived quantities come from the known base quantities. Base quantities are length, time, mass, electric current, temperature, substances, and luminosity. So a derived quantity uses these to form a different unit. Speed uses time and length units, for example.
Conversion of Units
Anything that we can measure comprises, a number, and the associated unit. Without the unit, the number means nothing. Sometimes there are multiple units involved. When this happens, we must convert to something meaningful to us. Lets convert 15 inches to centimeters. 1 in = 2.54 cm.
\[ 15 in * \frac{2.54 cm}{in} = 38.1 = 38 cm \]
You see that the inches cancelled out. This is the trick to knowing if you set it upright. The correct units will cancel out and it will leave you with the one unit that you want.
Example 1
We consider the age of the universe to be 14 billion years old. Let’s write this in scientific notation.
\[ 14,000,000,000 = 1.4 * 10^10 \]
Example 2
How many significant figures do the following numbers have? 214, 81.6, and .0086?
214 = 3 significant digits, 81.6 = 3 significant digits, and .0086 has 2 significant digits
Example 3
Write out the following numbers in scientific notation. 21.8, .0068
21.8 = 2.18 * 10^1 and .0068 = 6.8 8 10^3
Write out the following number. \( 8.69 * 10^4 \)
\[ 8.6900 \]
This section was about the basics of measurements and uncertainty. When measuring, you must always take into account the uncertainty of your tool. As you state your calculation, give the correct number of significant digits.
Vectors
Quantities like mass and temperature are described with a number called a scalar. Other quantities like displacement, velocity, or force have a direction associated with them and are called vectors. When looking at a vector on paper, a vector is an arrow with the length of the arrow representing the number and direction of the vector.
A vector can be described with a number and an angle, like \( A=19 \angle 25 \deg \). When doing basic math operations, you should use the component form to write vectors. So, if a line were placed with the tail of the vector at the origin of the coordinate system, x and y portions can be calculated like this:
\( y = 19 \sin 25^{\circ} = 8.02 \)
\( x = 19 \cos 25^{\circ} = 17.21 \)
This is how you calculate vectors.
If you have the x and y vectors, you can calculate the length of the corresponding line by:
\( \sqrt{x^{2} + y{2}} \)
\( \sqrt{8.02^{2} + 17.21^{2}} \)
\( \sqrt{64.32 + 296.18} \)
\( \sqrt{360.5} = 19 \)
To find the original angle we use:
\( \theta = \tan^{1} \frac{y}{x} \)
So:
\( \theta = \tan^{1} \frac{8.02}{17.21} \)
\( \theta = 25^{\circ} \)
Example 1
Write out the components of the vector \( c = 47 \angle 193^{\circ} \)
The x component is:
\( x = 47 \cos 193^{\circ} = 45.8 \)
The y component is:
\( y = 47 \sin 193^{\circ} = 10.6 \)
The use of unit vectors simplifies the mathematical operations on vectors. In two dimensions, unit vectors are vectors of unit value and in the +x and +y directions.
So the first vector would be written as \( 18.4i + 13.8j \) and the second vector is
\( 45.8i  10.6j \).
You can add vectors by adding the individual components. If you have:
\( (18.4i + 13.8j) + (45.8i  10.6j) \)
You just add the individual components together. The combined vector becomes:
\( 27.4i + 3.2j \)
Subtracting vectors can also be done. It is a bit quirky, however. You will use the opposite sign of the second vector.
\( (18.4i + 13.8j)  (45.8i  10.6j) \)
This gives the result of:
\( 64.2i + 24.4j \)
We should also know how to work backward when given the components. When we do this properly it will give us the length of the vector and its angle.
Let us use the last vector:
\( 64.2i + 24.4j \)
\( V = \sqrt{64.2^{2} + 24.4^{2}} = 68.7 \)
\( \theta = \tan^{1}\frac{24.4}{64.2} = 20.8^{\circ} \)
When put in vector form:
\( V = 68.7 \angle 20.8^{\circ} \)
Example 2
Add the vectors:
\(A=13 \angle 50^{\circ}\)
\(B=15 \angle 60^{\circ} \)
\(C=17 \angle 20^{\circ} \)
The first step is to put each vector in its component form.
\( A = \sin 50 * 13 = 10 = y \)
\( A = \cos 50 * 13 = 8.4 = x \)
\( A = 8.4i + 10.0j \)
\( B = \sin(60) * 15 = 13 = y \)
\( B = \cos(60) * 15 = 7.5 = x \)
\( B = 7.5i 13.0j \)
\( C = \sin 120 * 17 = 8.5 = y \)
\( C = \cos 120 * 17 = 14.7 = x \)
\( C = 14.7i 8.5 j \)
Sum all the i’s and all the j’s. You get:
\( 1.2i 11.5j \)
That is your x and y dimensions
Now use the Pythagorean theorem with those values
\( \sqrt{11.5^{2} + 1.2^{2}} = 11.6 \)
That gives you the length of the new vector
Now, let’s find the new angle
We use the inverse tangent to do this
\( \tan^{1} \frac{11.5}{1.2} = 84^{\circ} \)
So the new vector is:
\( 11.6 \angle 84^{\circ} \)
Use as many diagrams as possible when doing problems like this. Diagrams help keep your thought processes straight and help prevent silly errors.
Example 3
Find the resultant of the two forces \(800(N) \angle 47^{\circ}\) and \(600(N) \angle 140^{\circ}\)
I will write out all the calculation steps first, before I actually compute them
\(800 \cos 47^{\circ} = 545.6 \)
\(800 \sin 47^{\circ} = 585.1 \)
\(600 \cos 140^{\circ} = 460.0 \)
\(600 \sin 140^{\circ} = 385.7 \)
Vector 1 is: \(545.6i + 585.1j\)
Vector 2 is: \(460.0i + 385.7j\)
Add the vectors together
We get:
\(85.6i + 970.8j\)
Use Pythagorean theorem
\(\sqrt{85.6^{2} + 970.8^{2}}\)
The length of our new vector is:
\(= 974.6 \)
The new angle of our vector is:
\(\tan^{1}\frac{970.8}{85.6} = 85^{\circ} \)
So:
\(974.6(N) \angle 85^{\circ}\)
Scalars
The dot product of two vectors produces a scalar. Scalars, as you know, do not have a direction associated with them. There are two ways to compute a dot product.
\(A*B = AB \cos \theta\)
And
\(A*B = a_xb_x + a_yb_y\)
Example 4
Form the dot product of \(A = 23 \angle 37^{\circ}\) and \(B = 14 \angle 35^{\circ}\)
Use the first definition
\(A*B = A*B \cos \theta = 23*14 \cos 72^{\circ} = 100\)
Now use the second definition
This requires the components of both vectors
We learned how to do this earlier, it is the same process
We get:
\(18.4i + 13.8j\) and \(11.5i + (8.0)j\)
\(A*B=100\)
Cross Product
The cross product of A*B also produces a vector.
\(A*B \sin \theta\)
Example 5
Form the cross product of \(A = 23 \angle 37^{\circ}\) and \(B = 47 \angle 193^{\circ}\)
This is A*B* sine of the difference in the angles.
The difference in angles is 156 degrees.
\(A*B = 23*47*\sin 156^{\circ} = 440\)
We can also use determinants to get the same result
Calculate the components of each vector:
Cross multiply. This is why it is called the cross product
\(A*B = 18.4(10.6)  13.8(45.8) = 440\)
If it is a simple problem then use the first definition any time you can. Otherwise, use the second definition.
Kinematics
The motion of a particle is described by giving position, velocity, and acceleration, usually as a function of time. We start with one dimension. Distance means total distance traveled, while displacement is the actual difference between endpoint and the beginning.
\[Velocity = \frac{displacement}{time}\]
and:
\[Acceleration = \frac{v2v1}{time}\]
This velocity calculation fits our present definition and is equivalent to drawing a straight line between points on the smooth curve of x versus t. Calculating velocity this way presents a problem, because, depending on the interval, we will get different answers for the velocity. To find the velocity at t=1, we found x at t=1 and x at t=2 and performed the velocity calculation. This is not a good approximation to the velocity at t=1 because the average is between t=1 and t=2. For better approximations, shorten the time interval centered about t=1. Getting it as small as you can will give better results. So, the slope of the straight line tangent to the curve at x=1 represents the velocity at x=1.
Instantaneous Velocity and Acceleration
A more versatile definition of velocity is \(\frac{\Delta x}{\Delta t}\), where the interval \(\Delta t\) is very small and centered about the time where the velocity is desired. This approach leads to a general method for obtaining an expression for velocity that can be evaluated at any point rather than going through the numeric calculation whenever we want a velocity. This definition is stated as the instantaneous velocity.
\[v = \lim\limits_{\Delta t \to \0} = \frac{dx}{dt}\]
This definition of the derivative is the slope of the tangent to the curve evaluated at the point in question. If we want to find the velocity of any particle traveling according to a polynomial relationship between x and t, all we need is a general technique for finding the slope of the tangent to the polynomial at any point.
The instantaneous acceleration is defined as the value of \(\Delta v / \Delta t\) as \(\Delta t\) approaches zero.
\[a = \lim\limits_{\Delta t \to \0} = \frac{dy}{dt}\]
The instantaneous acceleration is the slope of the tangent to the curve of v versus t. The general expression for the slope of any polynomial of the form \(x = ct^{n}\). The expression for the slope at any point is \(cnt^{n1}\). Stating this in calculus terms, for any function \(x = ct^{n}\), the derivative of the function is \(cnt^{n1}\). This can be verified in the case of the parabola by taking successively smaller intervals of \(\Delta t\) and \(\Delta x\) at any point t to verify that the slope at any point is 2t.
Kinematic Equations of Motion
The kinematic equations of motion are derived under the assumption of constant acceleration. While this may seem at first to be a restriction, there are a large number of problems where the acceleration is constant. The simplest and most obvious are falling body problems, or problems involving bodies falling on or near the surface of the earth where the acceleration due to gravity is a constant.
Starting with the assumption of constant acceleration, we can write:
\[a = \frac{v_f  v_o}{t}\]
This can be rearranged to this form:
\[ta = v_f  v_o \longrightarrow ta + v_o = v_f \longrightarrow v_f = v_o + ta \]
We define the average velocity as:
\[v_{avg} = \frac{v+ v_o}{2}\]
We define displacement as:
\[x = x_o + v_{avg}t\]
We can now substitute \(v_{avg}\) with the previous equation.
\[x = x_o + \frac{v + v_ot}{2}\]
One more substitution for \(v = v_o+ at\) and we get:
\[x= x_o + v_ot + (½)at^2\]
We now have the four kinematic equations of motion.
\[v = v_o + at\]
\[x  x_o = (½)(v + v_o)t\]
\[x  x_o = v_ot + (½) at^2\]
\[v^2 = v_o^2 + 2a(x  x_o)\]
The first three equations relate displacement, velocity, and acceleration in terms of time, while the fourth equation does not contain the time.
Example 1
A train starts from rest and moves with constant acceleration. On first observation, the velocity is 20 m/s and 80 s later the velocity is 60 m/s. At 80 s, calculate the position, average velocity, and constant acceleration over the interval.
Calculate the acceleration:
\[a =\frac{v_fv_o}{t} = \frac{60\text{ m/s}  20\text{ m/s}}{80\text{ s}} = 0.50 \text{ m/s^2}\]
Calculate the distance traveled over this 80 s:
\[x = v_ot + (½)at^2 = (20\text{ m/s})(80\text{ s}) + (½)(0.50\text{ m/s^2})6400\text{ s^2}= 3200\text{ m}\]
The average velocity is:
\[v_{avg} = \frac{v_f+v_o}{2} = {20\text{ m/s}+60\text{ m/s}}{2} = 40\text{ m/s}\]
If the acceleration is constant, then the average velocity is the average of 20 m/s and 60 m/s, or 40 m/s, and at an average velocity of 40 m/s and 80 s, the distance traveled is 3200 m.
Example 2
Using the above problem, calculate the position of the train at 20 s.
\[x = v_ot + (½)at^2 = (20\text{ m/s})(20\text{ s}) + (½)(0.50\text{ m/s^2})(20\text{ s})^2 = 500 \text{ m}\]
Example 3
Using the above problem, find the time required for the train to reach 100m.
\[xx_o = v_ot+(½)at^2 \longrightarrow 100\text{ m} = (20\text{ m/s})t + (½)(0.50\text{ m/s^2})t^2\]
\[t = \frac{80{\pm} \sqrt{64004(1)(400)}}{2(1)} = \frac{80{\pm}89}{2} = 4.5, 85\]
The negative answer is not correct because we do not want negative time, so t=4.5s
Example 4
Using the above problem, find the velocity of the train at 120 m.
\[v^2 = v_o^2 + 2ax = (20\text{ m/s})^2 + 2(0.50\text{ m/s^2})120m = 400\text{ m^2/s^2} + 120\text{ m^2/s^2} = 520\text{ m^2/s^2} = 23\text{ m/s}\]
Example 5
Two vehicles are at position x=0 and t=0. Vehicle 1 is moving at constant velocity of \(30 \text{ m/s}\). Vehicle 2, starting from rest, has acceleration of \(10 \text{ m/s^2}\). Where do they pass?
Translated into algebra, this is asking what is the value of x when they pass? This can be determined by writing equations for the position of each vehicle and equates
\[ x_1 = v_10 t = 30 \text{ m/s} \text{ t} \]
and
\[x_2=(½)a_2t^2=(5\text{ m/s^2})t^2\]
Setting x1=x2 gives the time when they pass as \(30t=5t^2\) or \(5t(t6)=0\), so the vehicles pass at t=0 and t=6. Putting t=6 in either equation for x gives 180 m as the distance.
Example 6
In the above situation, when do the vehicles have the same velocity?
In algebra, this means to set the equations for velocity equal (v1=v2) and solve for the time. Remember that three of the four equations of motion are functions of time, so most questions are answered by first calculating the time for a certain condition to occur.
\[30\text{ m/s}=(10\text{ m/s^2})t = 3.0\text{ s}\]
Now that we know when, we can calculate where they have the same velocity. Use either equation for position and t=3.0s
\[x_1{t=3.0s}=(30\text{ m/s})(3.0s)=90\text{ m}\]
Example 7
For the situation in problem 5, what are the position, velocity, and acceleration of each vehicle when vehicle 2 has traveled twice the distance of vehicle 1?
The time when this occurs is when x2=2x1 so:
\[5t^2=60t\]
Or
\[5t(t12)=0\]
This gives the time of t=0 and t=12. The time t=0 is correct but t=12 is what we are looking for because we want to know something after some time has passed. At t=12,
\[x_1=(30\text{ m/s})(12\text{ s})=360\text{ m}\]
From the original statement of the problem,
\[v_1 0=30\text{ m/s}\]
and \(a_1=0\). Now solve for the remaining variables for the second vehicle by substitution.
\[x_2=(5\text{ m/s^2})(12\text{ s})=720\text{ m}\]
\[v_2=(10\text{ m/s^2})(12\text{ s})=120\text{ m/s}\]
From the original statement of the problem, \(a_2=10\text{ m/s}\).
Example 8
Two trains are traveling along a straight track, one behind the other. The first train is traveling at \(12\text{ m/s}\). The second train, approaching from the rear, is traveling at \(20\text{ m/s}\). When the second train is 200 m behind the first, the operator applies the brakes, producing a constant deceleration of \(0.20\text{ m/s^2}\). Will the trains collide, and if so, where and when?
First, diagram the situation. Write down the equations for each train and the information provided in the problem.
Train 1  Train 2 
\(x1=200m +(12\text{ m/s})t\)  \(x2=(20 \text{ m/s})t(0.10 \text{ m/s^2})t^2\) 
\(v1=12 \text{ m/s}\)  \(v2=20 \text{ m/s}(0.20 \text{ m/s^2})t\) 
\(a1=0\)  \(a2=0.20 \text{ m/s^2}\) 


Take t=0 when the brakes are applied and the first train is 200 m ahead of the second. This makes the position of the first train 200 m at t=0.
The question as to whether the trains collide means, is there a real time solution to the equation resulting from setting x1=x2?
\[200m+(12 \text{ m/s})t=(20 \text{ m/s})t(0.10 \text{ m/s^2})t^2\]
If there are no real solutions to the equation, then the trains do not collide. Drop the units, and write \(0.10t^28t+200=0\) which is solved by the quadratic formula. In a calculator, you can see there are no real solutions, so the trains never collide.
Example 9
Change the previous problem by giving the second train an initial velocity of 25 m/s. This will give a real time for the collision. Find the collision time.
Setting the expressions for x1 ands x2 equal and dropping the units produces
\[200+12t=25t0.10t^2\]
Or
\[0.10t^213t+200=0\]
This equals \(17.8, 112\)
Train 1  Train 2 
\(x_1=200 \text{ m}+(12 \text{ m/s})t\)  \(x_2=(25 \text{ m/s})t(0.10 \text{ m/s^2})t^2\) 
\(v_1=12 \text{ m/s}\)  \(v_2=25 \text{ m/s}(0.20 \text{ m/s^2})t\) 
\(a_1=0\)  \(a_2=0.20 \text{ m/s^2}\) 
The two times correspond to when x1=x2. The earliest time is the first coincidence and the end of the physical problem. The position at this time can be obtained from either expression for x.
\[x_1=200m+(12 \text{ m/s})17.8\text{ s}=414 \text{ m}\]
Verify this distance by using x_2. The velocity of the second train at collision is
\[v_2=25 \text{ m/s}(0.20 \text{ m/s^2})17.8\text{ s}=21.4 \text{ m/s}\]
The relative velocity between the two trains is \(v=1.4 \text{ m/s}\). The two times are the result of the quadratic in t. The two solutions occur when the curves cross. The equation for \(x_1=200+12t\) is a straight line of slope 12 starting at 200. The equation for \(x_2=25t0.10t^2\) is a parabola that opens down. While the mathematics produces two times, the reality of the problem dictates the earlier time as the one for the collision.
Falling Body Problems
The kinematic equations of motion for constant acceleration can be applied to a large collection of problems known as falling body problems. These problems are where the constant acceleration is the acceleration due to gravity on the surface of the Earth. I will list them here.
\[v_{avg} = \frac{v+v_{0}}{2}\]
\[v = v_{0} + at\]
\[x  x_{0} = (½)(v + v_{0})t\]
\[x  x_{0} = v_{0}t + (½)at^{2}\]
\[v^{2} = v_{o}^{2} + 2a(x  x_{o})\]
Example 1
Consider a ball dropped from the top of a 40 m tall building. Calculate everything possible.
You will not be able to calculate everything at first.
The trick is to calculate what you can and those answers will lead you to the rest of the calculations.
We know:
displacement=40 meters
\[x=0\text{ } v_{o}=0\text{ } a=9.8m/s^{2}\]
Since most of the kinematic equations contain the time, this is usually one of the first things to calculate.
Use the equation that has all the variables included in it.
\[xx_{0}=v^{2}_{0}t+(½)at^{2}\]
\[40m = (½)(9.8m/s^{2})t{2}\]
\[= 2.9s\]
Knowing the time, we can calculate the velocity using the second equation.
\[v = v_{0} + at\]
\[v = 0 + 9.8m/s^{2}(2.9)\]
\[= 28m/s\]
Example 2
Now, using the original problem, let us add an initial velocity of 8.0m/s when the ball is thrown down instead of just dropped. Find the time for the ball to reach the ground and the velocity on impact.
First, we calculate the time for the ball to hit the ground as before.
\[x  x_{0} = v_{0}t + (½)at^{2}\]
\[40m = 8.0t + (½)(9.8t^{2}\]
\[40m = 8.0t + 4.9t^{2}\]
\[4.9t^{2} + 8.0t  40m = 0\]
Use a quadratic formula.
\[t = \frac{80\pm\sqrt{644(4.9)(40)}}{2(4.9)}\]
\[t = 2.2s\]
The velocity at the ground level is:
\[v^{2} = (8.0m/s^{2})^{2} + 2(9.8m/s^{2})(40m) = 848m^{2}s^{2} = 29m/s\]
Example 3
Using the first problem, now throw the ball up from the top of the building with a velocity of 8.0 m/s. Find the time for the ball to reach the ground and the velocity on impact.
Take x as the displacement as positive down and the velocity as negative up. It is important to remember that the sign of the velocity is opposite that of the displacement. It does not matter whether the velocity is negative and the distance down is positive, only that they are opposite. Calculate the time of flight.
\[40m = (8.0m/s)t + (4.9m/s^{2})t^{2}\]
\[4.9t^{2}  8.0t  40 = 0\]
\[t = \frac{8.0\pm\sqrt{644(4.9)(40)}}{9.8} = 3.8s\]
The positive time is the correct choice.
The velocity when the ball strikes the ground is:
\[v^{2} = (8.0m/s^{2}) + 2(9.8m/s^{2})40m = 848m^{2}/s^{2} = 29m/s\]
Notice that whether the v term is a positive number or negative number, the result is the same. If the ball is thrown up with a certain velocity or down with the same velocity, the velocity at impact is the same. This is to be expected from the symmetry of the equations. If the ball is thrown up with a certain velocity, then on the way down it passes the same level with that same velocity.
Example 4
For the situation in the above problem, calculate the maximum height above the top of the building and the time for the ball to reach maximum height.
The time for the ball to reach maximum height is from equation 2. Note that at maximum height, the velocity must be zero. Watch the signs closely. It does not matter how you choose the signs, but the acceleration has to be opposite the velocity.
\[v = v_{0} + at\]
\[0 = 8.0m/s  (9.8m/s^{2})\]
\[t = 0.82s\]
Because of the symmetry, it takes the ball the same amount of time to reach maximum height as it does for the ball to return to the original level.
Calculate the height above the top of the building from equation 5.
\[v^{2} = v_{o}^{2} + 2a(x  x_{o})\]
\[0^{2} = (8.0m/s^{2})  2(9.8m/s^{2})(x  x_{0}) = 3.3m\]
Example 5
A bottle of champagne is dropped by a balloonist. The balloon is rising at a constant velocity of 3.0 m/s. It takes 8.0 s for the bottle of champagne to reach the ground. Find the height of the balloon when the bottle was dropped, the height of the balloon when the bottle reached the ground, and the velocity with which the bottle strikes the ground.
There are several possibilities concerning the origin and direction of the coordinate system. Take the origin at the height of the balloon when the bottle is dropped and the position of the balloon at t=0. Take the displacement as positive down. The main reason for taking the displacement as positive down is that the acceleration is down and the initial velocity is up, making two positives and one negative. As time goes on, displacement, velocity, and acceleration will be positive.
First, write the equation for the height of the balloon starting from the time when the bottle is dropped.
\[x = (3.0m/s)t + (4.9m/s^{2})t^{2} = (3.0m/s)(8.0s) + (4.9m/s^{2})(64s^{2}) = 290m\]
The height of the balloon when the bottle was dropped plus the amount the balloon rose in the 8.0s it took the bottle to reach the ground:
\[290m + (3.0m/s)8.0s = 314m\]
The velocity on impact :
\[v = v_{0} + at = 3.0m/s + (9.8m/s^{2})(8.0s) = 75m/s\]
Example 6
A parachutist descending at a constant rate of 2.0 m/s drops a smoke canister at a height of 300m. Find the time for the smoke canister to reach the ground and its velocity when it strikes the ground. Then find the time for the parachutist to reach the ground, the position of the parachutist when the smoke canister strikes the ground, and an expression for the distance between the smoke canister and the parachutist.
The time for the smoke canister to reach the ground is from equation 4.
\[x = v_{0}t + (½)at^{2}\]
\[300m = (2.0m/s)t + (4.9m/s^{2})t^{2}\]
Without units, the equation is \(4.9t^{2} + 2.0t  300 = 0\) with solutions.
\[t = \frac{2\pm\sqrt{44(4.9)(300)}}{2*4.9}\]
\[= \frac{2\pm76.7}{9.8} = 7.6\]
The time for the canister to hit the ground is 7.6s.
The velocity when it strikes the ground is:
\[v^{2} = (2.0m/s)^{2} + 2(9.8m/s^{2})300m = 5884m^{2}/s^{2}= 77m/s\]
The time for the parachutist to reach the ground is from equation 3.
\[300m = (2.0m/s)t = 150s\]
When the canister strikes the ground, the parachutist has dropped:
\[(2.0m/s)7.6 = 15m\]
And is 285 m above the ground. The expression for the distance between the canister and the parachutist is:
\[x_{c}  x_{p} = (2.0m/s)t + (4.9m/s^{2})t^{2}  (2.0m/s)t = (4.9m/s^{2})t^{2}\]
Example 7
A coconut is dropped from a height of 60m. One second later, a second coconut is thrown down with an initial velocity. Both coconuts reach the ground at the same time. What was the initial velocity of the second coconut?
In problems where there is a time delay, it is usually best to calculate the position, velocity, and acceleration of the first particle at the time when the second particle starts to move. It is possible to do time delay problems with a time differential in one set of equations. The difficulty with this approach is that it is easy to get an algebraic sign wrong. If you say that the time for the second particle is the time for the first particle plus the difference between them, then it is essential that the algebraic sign of the difference be correct. It is much easier to do them, in this slower but inherently more accurate way. First, calculate the state of the first particle when the second one begins moving. Then write the two sets of equations describing the motion with this instant as t=0.
Calculate the position and velocity of the first coconut at the end of 1s, the time when the second one starts.
\[x = (½)at^{2} = (½)(9.8m/s^{2})(1.0s)^{2} = 4.9m\]
\[v = at = 9.8m/s^{2}(1.0s) = 9.8m/s\]
Since position, velocity, and acceleration is positive down, orient the coordinate system for positive down with the origin at the top. At the instant the second coconut is thrown down, the first coconut has position 4.9m, velocity 9.8m/s and acceleration 9.8m/s^{2}.
Since both coconuts strike the ground at the same time, use the conditions of the first coconut to find the total time. First, calculate the velocity at impact of the first coconut.
\[v_{1}^{2} = v_{1o}^{2} + 2a(x_{1}x_{1o}) = (9.8m/s)^{2} + 2(9.8m/s^{2})(604.9)m = 1176m^{2}/s^{2} = 34.3m/s\]
The time for the second coconut to reach the ground is the same as the time for the first coconut to go from 4.9 m to 60 m or the time for the first coconut to go from 9.8 m/s to 34.3 m/s.
This comes from \(v = v_{0} + at\) where the velocity of the first coconut when the second one is thrown down, and v is the velocity of the first coconut at the ground.
\[34.3m/s = 9.8 m/s _ (9.8m/s^{2})t = 2.5s\]
Now that we have the time for the second coconut to travel the 60 m, we can find its initial velocity from \(xx_{0}=v_{2o}t + (½)at^{2}\) where t is the total time for the second coconut.
\[60m = v_{2o}(2.5s) + (4.9m/s^{2}(2.5s)^{2} =11.75m/s\]
Example 8
A boat is passing under a bridge. The deck of the boat is 15 m below the bridge. A small package is to be dropped from the bridge onto the deck of the boat when the boat is 25 m from just below the drop point. What boat speed is necessary to have the package land in the boat?
Calculate the time for the package to fall the 15.0 m using \(xx_{0} = v_{o}t+(½)at^{2}\)
\[15m = (4.9m/s^{2})t^{2} = 1.7s\]
The boat must move forward at 25 m/ 1.7s = 14 m/s.
Example 9
You are observing steel balls falling at a constant velocity in a liquid filled tank. The window you are using is 1m high, and the bottom of the window is 12m from the bottom of the tank. You observe a ball falling past the window taking 3.0s to pass the window. Calculate the time required to reach the bottom of the tank after the ball has reached the bottom of the window.
The observed velocity is \(1.0m/3.0s so 12 m = (1.0m/3.0st = 36s\)
Example 10
In a situation similar to the last example, the tank is filled with a different liquid, causing the acceleration in the tank to be \(6.0 m/s^{2}\) and the time to traverse the window .40 s. Calculate the height of the liquid above the window, the time to reach the bottom of the tank, and the velocity of the ball when it reaches the bottom of the tank.
From the data about the window, calculate the velocity of the ball at the top of the window.
\[xx_{o} = v_{t}t + (½)at^{2}\]
\[1.0m = v_{t}(.40s) + (3.0m/s^{2})(.40s)^{2} = 1.3m/s\]
The velocity at the bottom of the window is from:
\[v_{b}^{2} = v_{t}^{2} + 2a(xx_{o}) = (1.3m/s)^{2} + 2(6.0m/s^{2})1.0m = 3.7m/s\]
The time to reach the bottom of the tank is from \(xx_{o} = v_{b}t + (½)at^{2}\)
\[12m = (3.7m/s)t + (3.0m/s^{2})t^{2}\]
Eliminate the units
\[t = \frac{3.7\pm \sqrt{(3.7)^{2}  4(3.0)(12)}}{2(3)} = \frac{3.7\pm12.6}{6} = 1.5\]
This positive time is for the ball to travel from the bottom of the window to the bottom of the tank.
Assuming that the ball started at zero velocity, the distance from the top of the liquid to the top of the window will come from \(v_{t}^{2}=v_{o}^{2}+2ax_{t}\)
\[(1.3m/s)^{2} = 2(6.0m/s)x_{t} = .14m\]
The velocity with which the ball strikes the bottom of the tank is:
\[v = v_{b} + at = (3.7m/s) + (6.0m/s^{2})1.5s = 12.7m/s\]
Example 11
A ball is observed to pass a 1.4m tall window going up and later going down. The total observation time is .40s(.20s going up and .20s going down). How high does the ball rise above the window?
This is another example of a question that seems totally unrelated to the information in the problem. When you don’t know where to start and you do not see a clear path to the desired answer, simply start where you can, calculating what you can and hopefully learning enough to answer the specific question.
One of the first things we can calculate in this problem is the average velocity at the middle of the window.
\[v_{avg} = (1.4m/2.0s) = 7.0m/s\]
This average velocity is the velocity of the ball on the way up and on the way down at the middle of the window. Add this feature to the problem. With this information, we can use \(v^{2} = v_{o}^{2} + 2ax_{t}\) to find the distance the ball rises above the midpoint of the window.
\[(7.0m/s)^{2} = 2(9.8m/s^{2})x_{t} = 2.5m\]
So the ball rises 2.50.7=18 m above the top of the window.
View the ball as decelerating as it goes up past the window, ands find \(v_{b}\) at the bottom of the window from \(xx_{o}=v_{o}t+(½)at^{2}\).
\[1.4m = v_{b}(.20s)(4.9m/s^{2})(.20s)^{2} = 8.0m/s\]
Velocity and displacement are taken as positive, so acceleration is negative.
Now the distance to maximum height, where velocity is zero, is:
\[(8.0m/s)^{2} = 2(9.8m/s^{2})x_{m} = 3.2m\]
Again, the maximum height of the ball above the top of the window is 3.2 m  1.4 m = 1.8m.