Question

A boy throws a ball up in the air with a velocity of 9 m/s when they standing at the edge of a cliff 22 m high. If the ball falls and lands at the base of the cliff, what is the velocity of the ball when it lands?

Answer 1

The final velocity of the ball when it lands is approximately -65.6 m/s.

Step-by-step explanation:

To find the velocity of the ball when it lands, we need to use the principles of projectile motion. Since the ball is thrown up, its initial velocity is 9 m/s. We can use the equation v_f = v_i + 2ad to find the final velocity, where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and d is the displacement.

Since the ball is only affected by gravity, the acceleration is -9.8 m/s² (negative since it acts in the opposite direction of the initial velocity). The displacement is equal to the distance between the cliff and the ground, which is 22 m. Plugging in the values, we get:

v_f = 9 + 2*(-9.8)*22

Solving this equation, we find that the final velocity of the ball when it lands is approximately -65.6 m/s.