Published by:
CGP EDU Academic Team
Published on: September 13, 2026
A body A is projected upwards with a velocity of \(28 \; m \; i \; s\) . 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 \{t - 4\} sec will be the time of flight of the second body. Since \(h_1 \quad h_2\)
\(98t - \frac{1}{2}gt^{2} - 98(t - 4) - \frac{1}{2}g(t - 4)^{2}\)
On solving, we get t = 12 seconds
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