Question

A body is projected horizontally from the top of a tall tower. At a time of 3.5 s from the projection, the horizontal and vertical displacements of the body are equal. The velocity of projection of the body is (acceleration due to gravity = 10 ms\(^2\))

• A. 30 ms\(^{-1}\)
• B. 35 ms\(^{-1}\)
• C. 15 ms\(^{-1}\)
• D. 17.5 ms\(^{-1}\)

Hint:

In projectile motion, the horizontal and vertical displacements can be equal if the right initial velocity is used.

Answer 1

The Correct Option is D

The horizontal displacement and vertical displacement of the body are equal at a time of 3.5 seconds. Let \( v_0 \) be the velocity of projection. 
The equations of motion for horizontal and vertical displacements are given by: 
1. Horizontal displacement: \( x = v_0 t \) 2. Vertical displacement: \( y = \frac{1}{2} g t^2 \) Since the displacements are equal at \( t = 3.5 \) s: \[ v_0 \cdot 3.5 = \frac{1}{2} \cdot 10 \cdot (3.5)^2 \] 
Now solving for \( v_0 \): \[ v_0 \cdot 3.5 = \frac{1}{2} \cdot 10 \cdot 12.25 \] \[ v_0 \cdot 3.5 = 61.25 \] \[ v_0 = \frac{61.25}{3.5} = 17.5 \, {ms}^{-1} \] Thus, the velocity of projection is 17.5 ms\(^{-1}\). 
Final Answer: 17.5 ms\(^{-1}\).