Question

Two balls A and B are thrown with same speed u from the top of a tower. Ball A is thrown vertically upwards and the ball B is thrown vertically downwards. Choose the correct statement:-

• A. Ball B reaches the ground with greater velocity
• B. Ball A reaches the ground with greater velocity
• C. Both the balls reach the ground with same velocity
• D. Cannot be interpreted

Answer 1

Answer: (C)

Both the balls reach the ground with same velocity

Answer 2

Let the height of the tower be $h$.
Let the initial speed of both balls be $u$.
Let the acceleration due to gravity be $g$. We take the downward direction as positive.

For Ball A (thrown vertically upwards):
Initial velocity at the top of the tower, $u_A = -u$ (upwards is negative).
Displacement from the top of the tower to the ground, $s_A = +h$ (downwards is positive).
Acceleration, $a_A = +g$ (downwards).
Let $v_A$ be the final velocity of Ball A when it reaches the ground. Using the kinematic equation $v^2 = u^2 + 2as$:
$v_A^2 = (-u)^2 + 2(+g)(+h)$
$v_A^2 = u^2 + 2gh$

For Ball B (thrown vertically downwards):
Initial velocity at the top of the tower, $u_B = +u$ (downwards is positive).
Displacement from the top of the tower to the ground, $s_B = +h$ (downwards is positive).
Acceleration, $a_B = +g$ (downwards).
Let $v_B$ be the final velocity of Ball B when it reaches the ground. Using the kinematic equation $v^2 = u^2 + 2as$:
$v_B^2 = (+u)^2 + 2(+g)(+h)$
$v_B^2 = u^2 + 2gh$

Comparing the squares of the final velocities, we have $v_A^2 = v_B^2 = u^2 + 2gh$.
Since both balls are moving downwards when they reach the ground, their velocities are in the same direction.
The magnitude of the final velocity for both balls is $\sqrt{u^2 + 2gh}$.
Since both the magnitude and direction of the final velocity are the same, both balls reach the ground with the same velocity.