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CGP EDU Academic Team
Published on: September 12, 2026
A ball is dropped from top of a tower of 100 m height. Simultaneously another ball was thrown upward from bottom of the tower with a speed of 50 m/s ( \(g = 10 \mathrm{m/s^2}\) . They will cross each other after
Text Solution
Verified by ExpertsThe correct answer is:
B
\(h_1 - \frac{1}{2} g t^2\) , \(h_2 - 50t - \frac{1}{2}gt^2\)
Given \(h_1 + h_2 = 100\,m\) ⇒ ⇒ \(50 \rightarrow 100 \rightarrow t \quad 2 \text{ sec}\)
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