Question

A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity 25 ms\(^{-1}\). Then the distance from the top of the tower, at which the two balls meet is

• A. 68.4 m
• B. 48.4 m
• C. 18.4 m
• D. 28.4 m
• E. 78.4 m

Hint:

For two objects moving toward each other under the same acceleration (like gravity), the meeting time is always \( \text{Total Distance} / \text{Relative Velocity} \). Here, \( 100 / 25 = 4 \text{ s} \).

Answer 1

Answer: (E)

Concept: This is a problem involving relative motion under gravity. While both balls are accelerating at \( g \), their relative acceleration is zero, which simplifies the time calculation.
• Kinematic Equation: \( s = ut + \frac{1}{2} at^2 \).
• Meeting Condition: The sum of the distances traveled by both balls must equal the total height of the tower.

Step 1: Calculate the time of meeting.
Let \( h_1 \) be the distance dropped by the first ball and \( h_2 \) be the distance climbed by the second. \[ h_1 = \frac{1}{2} gt^2 \] \[ h_2 = ut - \frac{1}{2} gt^2 \] Total height \( H = h_1 + h_2 = \frac{1}{2} gt^2 + ut - \frac{1}{2} gt^2 = ut \). Given \( H = 100 \text{ m} \) and \( u = 25 \text{ ms}^{-1} \): \[ 100 = 25 \times t \implies t = 4 \text{ seconds} \]

Step 2: Calculate the distance from the top.
The distance from the top is simply the distance \( h_1 \) dropped by the first ball in 4 seconds. Using \( g = 9.8 \text{ ms}^{-2} \): \[ h_1 = \frac{1}{2} \times 9.8 \times (4)^2 \] \[ h_1 = 4.9 \times 16 = 78.4 \text{ m} \]