Question

A ball is released from the top of a tower of height \( h \) meters. It takes \( T \) seconds to reach the ground. What is the position of the ball above the ground in \( T/5 \) seconds?

• A. \( 25 \, \text{h m} \)
• B. \( \frac{h}{25} \, \text{m} \)
• C. \( 24 \, \text{h m} \)
• D. \( \frac{24h}{25} \, \text{m} \)

Hint:

In free fall problems, use the equation \( h = \frac{1}{2} g T^2 \) to calculate the distance fallen and subtract from the total height to find the position above the ground.

Answer 1

The Correct Option is D

The position of the ball is governed by the equation for free fall: \[ h = \frac{1}{2} g T^2 \] Where \( g \) is the acceleration due to gravity and \( T \) is the time it takes for the ball to reach the ground. We are asked to find the position of the ball above the ground after \( T/5 \) seconds. The distance traveled by the ball after \( T/5 \) seconds is: \[ h' = \frac{1}{2} g \left( \frac{T}{5} \right)^2 = \frac{1}{2} g \frac{T^2}{25} \] Thus, the height above the ground at \( T/5 \) seconds is: \[ \text{Height above the ground} = h - \frac{h}{25} = \frac{24h}{25} \] Therefore, the correct answer is \( \frac{24h}{25} \, \text{m} \).