Projectile Motion In Physics
These are my notes on projectile motion in physics.
In order to understand projectile motion, you have to look at the motion in two directions with one direction oriented in the direction of constant acceleration and the other direction at a right angle to it as to form an x-y coordinate system. In most problems where an initial velocity and angle with the horizontal is given, the velocity is written in component form.
\[v_y = v_o sin \Theta\]
\[v_x = v_o cos \Theta\]
Note that motion is effectively separated into horizontal and vertical components. One is in the direction of constant acceleration and one is at a right angle to the acceleration.
The motion is a parabola in this coordinate system. The symmetry of parabolas is helpful in understanding the motion.
Usually, the velocity of the projectile is written in component form. The acceleration in the horizontal direction is zero and in the vertical direction is due to gravity. With the velocity components and the acceleration in the direction of an axis, we can write the six equations describing acceleration, velocity, and position in the x and y directions.
These equations are based on the kinematic equations of motion for constant acceleration:
\[a = \text{ constant}, v = v_o + at, \text{and} s = s_o + v_ot + (1/2)at^2\]
The six equations are:
\[a_x = 0\]
\[v_x = v_o \cos\Theta\]
\[x = v_o(\cos\Theta)t\]
\[a_y = -g\]
\[v_y = v_o \sin\theta - gt\]
\[y = v_o(\sin\Theta)t - (1/2)gt^2\]
The equations for x and y can be considered parametric equations in time. Parametric equations such as \(x = f(t)\) are equations that can be combined to produce \(y = f(x)\) or \(x = f(x)\). To find the position of the particle in x-y or y as a function of x, solve one equation for t and substitute into the other.
In this case, solve x =...for t because this is the simplest choice.
\[t = \frac{x}{v_o \cos\Theta}\]
Substitute into y=...to obtain:
\[y = (\tan\Theta)x - \frac{g}{2v^2_o \cos^2\Theta}x^2\]
This is of the form \(y=-ax^2+bx=x(-ax+b)\), which is a parabola that opens down and intercepts the y-axis (makes y=0) at x=0 and x=b/a.
The range, or value of x when y=0, can be determined from the factored form of this equation.
\[y=x{\tan\Theta-\frac{g}{2v^2_o \cos^2\Theta}x}\]
which tells us that y=0 at x=0 and :
\[x=\frac{2v^2_o}{g} \cos^2\Theta \tan\Theta= \frac{2v^2_o}{g}\]
The maximum range occurs for an angle of \(45\deg\), corresponding to \(\sin2\Theta=1\).
The main point is that the motion is a parabola, and the properties of parabolas can be used in solving problems in projectile motion. That the maximum range occurs at \(45\deg\) is not surprising and is generally not of interest in problems. The expression for maximum range is only of passing interest because the range is one of the easier things to calculate in a problem.
Do not consume precious memory space memorizing formulas for the range or time of flight. If you work problems by first writing down the six equations describing the motion, then the time of flight, range, and many other things are easily calculated.
In order to set up a problem in projectile motion, first orient one axis of a right angle coordinate system in the direction of constant acceleration. Remember to place the origin of the coordinate system and the positive direction for x and y for your convenience in solving the problems. Then write equations for acceleration, velocity, and position for each direction. Along with the symmetry for the motion, these equations can be used to find the characteristics of the projectile at any point in space or time.
Example 1
A soccer player kicks a ball at an initial velocity of 18 m/s at an angle of 36 degrees to the horizontal. Find the time of flight, range, max height, and velocity components at t=0, midrange, and impact.
The acceleration is down, so set up the coordinate system with x horizontal and y vertical, and place \(v_o\) on the graph. Write the velocity in component form, and calculate \(v_{xo}\) and \(v_{yo}\).
\[v_o=18 m/s\]
\[v_{yo}=v_o \sin 36 = 10.6 m/s \]
\[v_{xo}=v_o \cos 36 = 14.6 m/s \]
Now, write down the six equations governing the motion starting with the accelerations.
\[ a_x=0 \text{and }a_y=-9.8m/s^2 \]
\[v_x=14.6m/s \text{and } v_y=10.6m/s - (9.8m/s^2)t \]
\[ x=(14.6m/s)t^2 \text{and } y=(10.6m/s)t - (4.9m/s^2)t^2 \]
Now look at the motion, which is parabolic, keeping in mind the properties of parabolas.
The ball is on the ground at t=0 and t=t, the time of flight.
To find these times, set y=0
\[10.6t - 4.9t^2 = 0 \]
This gives \( t=0\) and \( t=(10.6/4.9)=2.16s\), the time of flight.
The range is the value of x at 2.16s.
\[x_{t=2.16} = (14.6m/s)2.16s = 31.5m \]
The max height and \(v_y=0\) occur at 1.08s. Therefore, the max height is y at t=1.08s.
\[y_{max} = y_{t=1.08} = (10.6m/s)1.08s - (4.9m/s^2)(1.08s)^2 = 5.73m \]
The velocity components are:
t=0: v(x)=14.6 m/s v(y)=10.6 m/s
t=1.08s v(x)=14.6 m/s v(y)=0
t=2.16s v(x)=14.6 m/s v(y)=-10.6 m/s
The procedure for doing the problem is to write the initial velocity in component form, then write the six equations for acceleration, velocity and position. Then perform the mathematical operations answering questions about the problem.
Example 2
for the situation in example 1, suppose that the field is covered with fog down to 3.0 m above the ground. What are the times for the ball's entry into and exit out of the fog?
The times for the ball's entry into and exit out of the fog, are found from the equation for y as a function of time with y set equal to 3.0 m.
\[3.0m = (10.6m/s)t - (4.9m/s^2)t^2 \]
This is a quadratic equation, and after removing the units, it reads \(4.9t^2 - 10.6t + 3.0 = 0\) and is solved with the quadratic formula.
\[\frac{10.6\pm \sqrt{10.6^2-4(4.9)3.0}}{2(4.9)} = \frac{10.6\pm7.3}{9.8} = 0.34,1.8 \]
Example 3
An airplane traveling at 100 m/s drops a bomb from a height of 1500m. Find the time of flight, distance traveled, and velocity components as the bomb strikes the ground.
Start by placing the origin of the coordinate system at the point where the bomb is released. Take the direction in which the bomb falls due to gravity as positive x and the horizontal position as y. This is different from the conventional orientation, but it is convenient in this problem because the six equations all come out positive.
Now write down the six equations governing motion.
\(a_x = 9.8m/s^2\) and \(a_y = 0 \)
\(v_x = (9.8 m/s^2)t\) and \(v_y = v_o = 100m/s\)
\(x = (4.9m/s^2)t^2\) and \(y = (100m/s)t\)
With these six equations we can answer all the questions in the problem. We need only translate the word questions into algebra questions.
A. The time for the bomb to reach the Earth means "Find the time when x=1500m".
\(1500m = (4.9m/s^2)t^2 = t = 17s\)
B. How far does the bomb travel horizontally means "Find the value of y when t=17s.
\(y_{t=17s} = (100m/s)17s = 1700m\)
C. Find the velocity components at impact means "Find v(x) and v(y) at t=17s.
\( v_{x,t=17} = (9.8m/s^2)17s = 167 m/s \)
\( v_{y,t=17} = (100m/s)\)
D. Where is the airplane when the bomb strikes the Earth means "What is y at 17s?
Remember that the plane and the bomb have the same velocity in the horizontal direction.
\(y_{plane} = (100m/s)17s = 1700m\)
Example 4
A baseball is hit at a 45 degree angle and a height of 0.90 m. The ball travels a total distance of 120m. What is the initial velocity of the ball?
What is the height of the ball above a 3.0 m fence from where the ball was hit?
This problem is unique in that it requests the initial velocity, a number usually given in the problem. Also note that the angle is 45 degrees, the angle for max range. The other interesting feature of this problem is the question concerning the height at some specific point down range.
Set up the problem in the conventional way assigning \(v_o\) to the initial velocity.
\(v_o \sin 45\deg\)
\(v_o \cos 45\deg\)
The next questions is how to handle the ball being hit at 0.90 m above the ground. Since \(v_o\) is specified at the point where the ball is hit, and this is where we start counting time, the origin for the coordinate system should be put here. We just need to keep in mind that this origin is 0.90 m above the ground level.
Now write the six equations.
\(a_x = 0\) and \(a_y = -9.8m/s^2\)
\(v_x = v_o \cos 45\deg\) and \(v_y = v_o \sin 45\deg - (9.8m/s^2)t\)
\(x = v_o (\cos 45\deg)t\) and \(y = v_o \sin 45\deg t - (4.9m/s^2)t^2\)
The first question reduces to finding \(v_o\) for the 120 m hit. Algebraically, this means that when \(y=-0.9m, x=120m\), so write the two equations with these conditions.
\(120m = v_o \cos 45\deg t\) and \(-0.9m = v_o(\sin 45\deg)t - (4.9 m/s^2)t^2\)
Note that \(v_o(\cos 45\deg)t = v_o(\sin 45\deg)t = 120m\)
\(0 = 0.90m + 120m - (4.9m/s^2)t^2\) and \(t=4.97s\)
This is the time of flight, so put this time into the equation for x.
\(120m = v_o(\cos 45\deg)(4.97s)\) to find \(v_o\) as \(v_o = \frac{120m}{\cos 45\deg*4.97s} = 34.1m/s\)
The next part of the problem asks for the height above the fence at 100m. Again, we need to find the time for x to be 100m down range and substitute this value of t into the y equation.
\(100m = (34.1m/s)(\cos 45\deg)t = 4.15s\)
This is the time for the ball to go 100m. The height of the ball at this time is:
\(y_{t=4.15s} = (34.1m/s)(\cos 45\deg)(4.15s) - (4.9m/s^2)(4.15s)^2 = 15.6m\)
Remember that this is 15.6m above the zero of the coordinate system, which is 0.90m above the ground, so the ball is 16.5m above the ground at this point and, for a 3m high fence, is 13.5 m above the fence.
Example 5
A physicist turned motorcycle stunt rider will jump a 20 m wide row of cars. The launch ramp is 30 degrees and 9.0 m high. The land ramp is also 30 degrees and is 6.0 m high. Find the minimum speed for the launch.
Diagram the problem
First, write down the six equations of motion.
\(a_x = 0\) and \(a_y = -9.8m/s^2\)
\(v_x = v_o \cos 30\deg\) and \(v_y = v_o \sin 30\deg - (9.8m/s^2)t\)
\(x = v_o(\cos 30\deg)t\) and \(y = v_o(\sin 30\deg)t - (4.9m/s^2)t^2\)
The minimum \(v_o\) is dictated by the condition that the rider be at x=20m and y=-3.0m.
So put these conditions into the equations for x and y.
\(20m = v_o(\cos 30\deg)t\) and \(-3.0m = v_o(\sin 30\deg)t - (4.9m/s^2)t^2\)
At this point we can solve for \(v_o\) by substituting \(t=20m/v_o \cos 30\deg\) from the first equation into the second equation.
\(-3.0m = (20m) \tan 30\deg - \frac{4.9*400m^3}{v^2_o(\cos^230\deg)s^2} = v_o=13.4m/s\)
This is the minimum velocity to make the jump.