Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A body A is projected upwards with a velocity of \(98\,m/s\) . The second body B is projected upwards with the same initial velocity but after 4 sec . Both the bodies will meet after
Text Solution
Verified by ExpertsThe correct answer is:
D
Let t be the time of flight of the first body after meeting, then \(\left(t-4\right)\) sec will be the time of flight of the second body. Since h_1 = h_2
\(98t - \frac{1}{2}gt^{2} = 98(t - 4) - \frac{1}{2}g(t - 4)^{2}\)
On solving, we get t = 12 seconds
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