Two bodies are thrown simultaneously from a tower with same…

Physics Motion in a Straight Line Mix Single Correct MCQ

Published on: September 12, 2026

Two bodies are thrown simultaneously from a tower with same initial velocity \(\mathbf{V_0}:\) one vertically upwards, the other vertically downwards. The distance between the two bodies after time t is

\(2v_0 t + \frac{1}{2} g t^2\)

2v_{0}t

\(v_0 t + \frac{1}{2} g t^2\)

V_{0}t

Text Solution

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The correct answer is:

For vertically upward motion, \(h_1 = v_0 t - \frac{1}{2} g t^2\) and for vertically down ward motion, \(h_2 = v_0 t + \frac{1}{2} g t^2\)

Total distance covered in t sec \(h = h_{1} + h_{2} = 2 \sqrt{v_{0} t}\) .

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