A student at the top of a building of height $h$ throws one ball upward with an initial velocity $v_{i}$ and then throws a second ball downwards with the same initial speed. How were the final velocities of the ball to compare when they reach the ground?

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Solution

Explanation:
1. When the ball is thrown upward
It has displacement, $s = - h$ ; Initial velocity; $u = v_{i}$ ; acceleration, $a = - g$
We have $2 a s = v^{2} - u^{2}$ (equation of motion)
Putting the values in the above equation: $- 2 g (- h) = v^{2} - {v_{i}}^{2} \Rightarrow v = \sqrt{{v_{i}}^{2} + 2 g h} \to (1)$
2. When the ball is thrown downward
It has displacement; $s = - h$ ; Initial velocity, $u = - v_{i}$ acceleration $a = - g$
We have $2 a s = v^{2} - u^{2}$ (equation of motion)
Putting the values in the above equation: $2 g h = v^{2} - {v_{i}}^{2} \Rightarrow v = \sqrt{{v_{i}}^{2} + 2 g h} \to (2)$
From (1) and (2) we can conclude that both the balls have the same final velocities. Hence, the final velocities of the ball are equal when they reach the ground.

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A student at the top of a building of height h throws one ball upward with an initial velocity viand then throws a second ball downwards with the same initial speed. How were the final velocities of the ball to compare when they reach the ground?

A student at the top of a building of height $h$ throws one ball upward with an initial velocity $v_{i}$ and then throws a second ball downwards with the same initial speed. How were the final velocities of the ball to compare when they reach the ground?