Question

A student throws a baseball horizontally at 25 meters per second from a cliff 45 meters above the level ground. Approximately how far from the base of the cliff does the ball hit the ground? [neglect air resistance.]

Answer 1

A baseball is thrown horizontally at 25 m/s from a cliff 45 m above the level ground and it hits the ground 75 m from the base of the cliff.

First, we will calculate the time the baseball is in the air by analyzing the vertical movement.

This is a uniformly accelerated motion, where the initial vertical velocity (u) is 0, the displacement (s) is 45 m and the acceleration is that of gravity (a = 9.8 m/s²).

We can calculate the time (t) in the air using the following kinematic equation.

$s = ut + \frac{1}{2} a t^{2} \\\\45m = (0m/s)t + \frac{1}{2} (9.8m/s^{2} ) t^{2} \\\\t = 3.0 s$

Horizontally, it travels for 3.0 s at a speed of 25 m/s. The horizontal distance traveled is:

$3.0s \times \frac{25m}{s} = 75 m$

A baseball is thrown horizontally at 25 m/s from a cliff 45 m above the level ground and it hits the ground 75 m from the base of the cliff.

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Answer 2

Neglecting air resistance, the distance from the base of the cliff where the baseball hit the ground is approximately 75.75m.

Given the data in the question;

• Initially, the baseball was at rest before it was thrown

• Initial velocity; $u = 0m/s$

• Final velocity; $v = 25m/s$

• Height of the cliff; $s = 45m$

First, we determine the time taken. Using the second equation of motion:

$s = ut + \frac{1}{2}gt^2$

Where s is the height of the cliff, u is initial velocity, t is time taken, a is acceleration due to gravity( $9.8m/s^2$)

We substitute in our values

$45m = 0 + \frac{1}{2}*9.8m/s^2\ *\ t^2\\\\t = \sqrt{ \frac{45m}{4.9m/s^2}}\\\\t = 3.03s$

Now, to get the horizontal distance, we simply multiply the velocity of the baseball and time taken

Horizontal distance = $25m/s\ *\ 3.03s$

Horizontal distance = $75.75m$

Therefore, neglecting air resistance, the distance from the base of the cliff where the baseball hit the ground is approximately 75.75m.

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