Projectile Motion

To understand this idea more, consider two marbles of the same size and weight. You release one marble from a specific height and throw the other horizontally from the same height. As long as you disregard wind resistance, both marbles will hit the ground at the same time because the horizontal component does not influence the vertical motion of the marble.

Suppose you have a body that rolls off a cliff with a velocity of 5 m/s. The body hits the ground at a distance d away from the base of a cliff that has a height of 30 m. Figure 3 shows the projectile motion without an angle, i.e. launched parallel to the horizontal. Calculate the range d covered by the object.

Solution

To calculate d, the distance from the base of the cliff, we need to understand more about the motion in x and y directions.

Assuming there is no air resistance and just the gravitational force acting on the ball, the speed in the x-direction will be 5m/s until the ball hits the ground. In the y-direction, the ball has a constant acceleration of 9.81m/s², which is caused by the gravitational force.

But what’s the initial velocity in the y-direction?

As mentioned previously, because the motion in both x and y directions are independent of each other, the 5m/s velocity in the x-direction has no impact on the movement in the y-direction. Hence, the ball rolls off the cliff with an initial velocity of 0m/s in the y-direction.

The displacement horizontally will be -30m because the downward direction is considered negative along with the acceleration of free fall, which is -9.81 m/s².

For the x-direction:

Initial velocity: u_x = 5 m/s

Distance traveled in the x-direction: d_x = ?

For the y-direction:

Initial velocity: u_y = 0 m/s

Displacement: s_y = -30 m

Acceleration due to free fall = a_y = -9.81 m/s²

From the motion in the y-direction, we can calculate the time t because time is the same in the x and y-direction. By using the second equation of motion and plugging in the values, we get:

\[s_y = u_y \cdot t + \frac{1}{2} a \cdot t^2\]

\[-30 m = 0 \cdot t + \frac{1}{2} (9.81 m/s^2) \cdot t^2\]

\[t = 2.47 s\]

Therefore, the time taken for the ball to reach the ground from a height of 30m is 2.47s.

To calculate the distance travelled from the base of the cliff d_x, we use the second equation of motion again, but this time it will be for the movement in the x-direction.

\[d_x = u_x \cdot t + \frac{1}{2} a_x \cdot t^2\]

\[d_x =5 m/s \cdot (2.47 s) + \frac{1}{2} (0) \cdot (2.47s)^2\]

\[d_x = 12.35 m\]

The distance the ball travels in the x-direction with an initial velocity of 5m/s from a height of 30m is 12.35m.

You can use any equation of motion to calculate a specific entity depending on the problem related to a projectile motion.

Take a look at the figure below. A cannonball is fired off a cliff at an initial velocity of 90m/s from a height of 25m from the ground at an angle of 53°. Calculate the distance the cannonball travels in the x-direction.

Solution

As you can see in the figure above, the ground is elevated 9m from the base of the cliff where the cannonball will land. This means that the displacement in the y-direction will not be 25m – it will be different.

First, let us resolve the velocity vector into its components.

Initial speed in the x-direction: V_x=90cos53m/s

Initial speed in the y-direction: V_y=90sin53m/s

The displacement will be -16m as the direction downward is taken as negative.

Using the second equation of motion and plugging in the values, we can calculate the time t it takes the cannonball to hit the ground from launch. Note that the time taken will be the same in either x- or y-direction.

\[s_y = u_y \cdot t + \frac{1}{2} a_y \cdot t^2\]

\[-16 = (90 \sin 53 m/s) \cdot t + \frac{1}{2} (-9.81 m/s^2) \cdot t^2\]

\[t = 14.45 s\]

As the air resistance is negligible, the speed in the x-direction will be consistent, i.e. it will be 90cos53 throughout the motion. We can figure out the distance travelled by multiplying speed in the x-direction with the time taken. Therefore,

\[d_x = (90 \cos 53 m/s) \cdot 14.45 s\]

\[d_x =

So, the horizontal distance the fired cannonball travels is 782.66m.

Suppose you have two objects at different heights, and you want to throw each of them in such a way that both objects should cover the same distance. There is no air resistance, both objects are of the same size and mass, and both are launched at the same speed.

The object thrown from a greater height will need to be launched at a decreasing angle and vice versa for the object thrown from a lower height. Hence, there exists a relationship between the height of release and the distance covered horizontally.