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JEE Main 2021 (25 Feb, Shift 2) · Motion Under Gravity

Motion Under Gravity question from JEE Main 2021 (25 Feb, Shift 2)

Q33JEE Main 2021 · 25 FEB · SHIFT 2Medium

Motion Under Gravity

A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25 m below the top. Both stones reach the bottom of building simultaneously. The height of the building is:

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Answer: (c)

Stone 1 after falling 5 m: $v = \sqrt{2 \times 10 \times 5} = 10$ m/s (take $g = 10$ m/s $^2$ ).
For the remaining time $t$ : stone 1 covers $H - 5 = 10t + 5t^2$ ; stone 2 covers $H - 25 = 5t^2$
Subtract: $20 = 10t \Rightarrow t = 2$ s
$H = 25 + 5(2)^2 = 45$ m. So the answer is (c) 45 m.

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