Definition of Projectile Motion: A projectile is anybody that is thrown or projected. In other words, a projectile is an object upon which the only force acting is gravity. The projectile motion has two elements: the horizontal projectile motion and the vertical projectile motion. The path of a projectile object and the projectile motion is captured and illustrated in the diagram below.
Projectile motion refers to the motion of an object projected into the air at an angle. A few examples of this include a soccer ball begin kicked, a baseball begin thrown, or an athlete long jumping. Even fireworks and water fountains are examples of projectile motion.
Path of a projectile showing projectile motion
Consider a projectile launched from 0 (in the projectile motion diagram above) with the initial velocity of magnitude u in a direction at angle A above the horizontal. 0, u, and A are shown in the projectile motion diagram above. These three variables are often present in projectile motion problems in physics and are used to derive the projection motion equation.
Deriving formulas of projectile motion
Let x denote the horizontal projectile motion or the horizontal component of displacement of the projectile motion.
Let y denote the vertical projectile motion or the vertical component of displacement of the projectile motion.
The horizontal projectile motion is analyzed separately from the vertical projectile motion.
| Horizontal projectile motion | Vertical projectile motion |
| Initial velocity = u cos A | Initial velocity = u sin A |
| Acceleration = 0 | Acceleration = – g (g = gravity) |
| Time taken = t | Time taken = t |
| Distance moved = x | Distance moved = y |
| x = ut cos A | y = (ut sin A) – ½ gt2 |
| Final speed vx = u cos A | Final speed vy = (u sin A) – gt |
The horizontal velocity vx remains constant but the vertical velocity vy decreases steadily from its initial value at a rate of g (gravity)
Total time taken is t = 2v0 sin q/g
Maximum Height is H = v0 sin2 q/2g
Time of falling equals time of rising i.e. tf = tr; tr = v0y/g, distance travelled along the y-axis y(t) = v0y – gt2/2, velocity along the y-axis is vy(t) = v0y – gt.
What follows is a general solution for the two dimensional motion of an object thrown in a gravitational field. This is usually termed a projectile motion problem. The thrown object is called the projectile. Its path is called the trajectory. We will answer all the usual questions that arise in a first year physics class regarding this motion. We will not consider air resistance. Without air resistance, the projectile will follow a parabolic trajectory. We will be throwing the projectile on level ground on planet Earth. It will leave the point of release, arc through the air along a path shaped like a parabola, and then hit ground a certain distance from where it was thrown.
As mentioned above, this is a two dimensional problem. Therefore, we will consider x and y directed displacements, velocities, and accelerations. The projectile will accelerate under the influence of gravity, so its y acceleration will be downward, or negative, and will be equal in size to the acceleration due to gravity on Earth. There will be no acceleration in the x direction since the force of gravity does not act along this axis.
On Earth the acceleration due to gravity is 9.8 m/s2directed downward. So, for this presentation acceleration in the y direction, or ay, will be -9.8 m/s2, and acceleration in the x direction, or ax, will be 0.0 m/s2.
Given the original conditions with which the projectile is thrown we will proceed to find the components of the original velocity and then move on to answer the following questions:
- How much time passes till the projectile is at the top of its flight?
- How high does the projectile rise?
- How much time passes till the projectile strikes the ground?
- How far away does the projectile land from its starting point?
Original or initial conditions
The original conditions are the size of the velocity and the angle above the horizontal with which the projectile is thrown.
General
Original size of velocity: vo, Original angle: q
Example
vo= 40.0 m/s
q = 35 degrees
Components of original velocity:
The usual first step in this investigation is to find the x and y components for the original velocity.
General
X component of original velocity: vox= vocos(q)
Y component of original velocity: voy= vosin(q)
Example
In the x direction: vox= vocos(q)
vox= (40.0 m/s)(cos(350))
vox= (40.0)(0.8191)
vox= 32.76
vox= 32.8 m/s
In the y direction: voy= vosin(q)
voy= (40.0 m/s)(sin(350))
voy= (40.0)(0.5735)
voy= 22.94
voy= 22.9 m/s
How much time passes until the projectile is at the top of its trajectory?
At the top of the trajectory the y, or upward, velocity of the projectile will be 0.0 m/s. The object is still moving at this moment, but its velocity is purely horizontal. At the top it is not moving up or down, only across.
Notice that the object is still in motion at the top of the trajectory; however, its velocity is completely horizontal. It has stopped going up and is about to begin going down. Therefore, its y velocity is 0.0 m/s.
We need to find out how much time passes from the time of the throw until the time when the y velocity of the projectile becomes 0.0 m/s. This y velocity at the top of the trajectory can be thought of as the final y velocity for the projectile for the portion of its flight that starts at the throw and ends at the top of the trajectory.
We will call this amount of time ‘the half time of flight’, since the projectile will spend one half of its time of flight rising to the top of its trajectory. It will spend the second half of its time of flight moving downward.
General
We can use the following kinematics equation: vf= vo+ at
Subscript it for y:
vfy= voy+ ayt
Solve it for t:
t = (vfy– voy) / ay
Plug in 0.0 m/s for vfy:
t = (0.0 m/s – voy) / ay
If the original y velocity and the y acceleration, i. e., the acceleration due to gravity, are plugged into the above equation, it will solve for the amount of time that passes from the moment of release to the moment when the projectile is at the top of its flight.
Example
Start with: t = (vfy– voy) / ay
Plug in 0.0 m/s for vfy:
t = (0.0 m/s – voy) / ay
Plug in values for voy and ay:
t = (0.0 m/s – 22.9 m/s) / – 9.8 m/s2
t = -22.9 / -9.8
t = 2.3s
In this example 2.3s of time passes while the projectile is rising to the top of the trajectory.
How high does the projectile rise?
Here you need to find the displacement in the y direction at the time when the projectile is at the top of its flight. We have just found the time at which the projectile is at the top of its flight. If we plug this time into a kinematics formula that will return the displacement, then we will know how high above ground the projectile is at when it is at the top of its trajectory.
General
Here is the displacement formula: d = vot + 0.5at2
We must think of this displacement in the y direction, so we will subscript this formula for y:
dy= voyt + 0.5ayt2
If now we plug in the half time of flight, which was found above, we will solve for the height of the trajectory, since the projectile is at its maximum height at this time.
Example
Starting with: dy= voyt + 0.5ayt2
Then plugging in known values:
dy= (22.9 m/s)(2.33 s) + (0.5)(-9.8 m/s2)(2.33 s)2
dy= 53.35 – 26.60
dy= 26.75
dy= 27 m
How much time passes until the projectile strikes the ground?
General
With no air resistance, the projectile will spend an equal amount of time rising to the top of its projectile as it spends falling from the top to the ground. Since we have already found the half time of flight, we need only to double that value to get the total time of flight.
Example
t = 2(2.33 s) = 4.66 = 4.7 s
This is the total time of flight.
How far away does the projectile land from its starting point?
The distance from the starting point on the ground to the landing point on the ground is called the range of the trajectory. This range is a displacement in the x direction. It is governed by the x velocity of the projectile. This x velocity does not change during the flight of the projectile. That is, whatever is the value of the x velocity at the start of the trajectory will be the value of the x velocity throughout the flight of the projectile. The x velocity remains constant because there are no accelerations in the x direction. The only acceleration is in the y direction, and this is due to the vertical pull of gravity. Gravity does not pull horizontally. Therefore, the calculation for the range is simplified.
General
Let us start with the general displacement formula:
d = vot + 0.5at2
Since we are working in the x direction, we should subscript this equation for x:
dx= voxt + 0.5axt2
Now, since the acceleration in the x direction is 0.0 m/s2, the second term in the above equation drops out, and we are left with dx= voxt
The velocity in the x direction does not change. The projectile maintains its original x velocity throughout its entire flight. So, the original x velocity is the only x velocity the projectile will have. We could, therefore, think of the last equation as dx= vxt
If we plug in the original x velocity for vx and the total time of flight for t, we will solve for the horizontal displacement, or range, of the trajectory.
Example
As shown in the general section above, start with dx= vxt
Plug in values. Remember that the x velocity is constant and always equal to its original value and that the time here is the total time of flight.
dx= (32.8 m/s)(4.66 s)
dx= 152.84
dx= 150 m
EXERCISES
Lets see how much you’ve learnt, attach the following answers to the comment below
A golfer practicing on a range with an elevated tee 4.9 m above fairway is able to strike a ball so that it leaves the club with a horizontal velocity of 20 m s–1. (Assume the acceleration due to gravity is 9.80 m s–2, and the effects of air resistance may be ignored unless otherwise stated.
- How long after the ball leaves the club will it land on the fairway? A. 2.0 s B. 1.0 s C. 3.0 s D. 4.0 s
- What horizontal distance will the ball travel before striking the fairway? A. 20 m B. 30 m C. 40 m D. 50 m
- What is the acceleration of the ball 0.5 s after being hit? 14.6 ms-2 B. 9.8 ms-2 C. 12 ms-2 D. 8.9 ms-2
- Calculate the speed of the ball 0.80s after it leaves the club. 21.5 ms-1 B. 23 ms-1 C. 34 ms-1 D. 42 ms-1
- With what speed will the ball strike the ground? 25 ms-1 B. 27.3 ms-1 C. 26 ms-1 D. 22.3 m s–1
