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JEE Main 2022 · 29 Jul · Shift 1, Physics question 32

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height h3\dfrac h3 while going up and coming down respectively.

Choose one option
Answer and worked solution

Official answer: B

Times. With T=2h/gT=\sqrt{2h/g} to the top, falling the remaining 2h3\frac{2h}3 takes T23T\sqrt{\frac23}. So t1=T(1−23)t_1=T\left(1-\sqrt{\tfrac23}\right), t2=T(1+23)t_2=T\left(1+\sqrt{\tfrac23}\right).

Ratio 3−23+2\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}.

(B)

Topics:Motion in a straight line
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