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If a ball is dropped from a 100 m cliff, it accelerates at a constant rate of 9.8 m/s2. Assuming its initial velocity is 0 m/s, its initial position is 0 m and its final position is 100 m away from the initial position, what is the velocity of the ball when it hits the ground?

Question

If a ball is dropped from a 100 m cliff, it accelerates at a constant rate of 9.8 m/s2. Assuming its initial velocity is 0 m/s, its initial position is 0 m and its final position is 100 m away from the initial position, what is the velocity of the ball when it hits the ground?

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Solution

To find the final velocity of the ball when it hits the ground, we can use the equation of motion:

v² = u² + 2as

where: v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

Given in the problem: u = 0 m/s (initial velocity), a = 9.8 m/s² (acceleration due to gravity), and s = 100 m (height of the cliff).

Substituting these values into the equation, we get:

v² = 0 + 29.8100 v² = 1960 v = sqrt(1960) v = 44.27 m/s

So, the velocity of the ball when it hits the ground is approximately 44.27 m/s.

This problem has been solved

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What is the velocity of the ball when it hits the ground?

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