Q 11-02-017NEETJEE MainEAMCET 2008 (Engineering)Medium
A body is thrown vertically up to reach its maximum height in $t$ seconds. The total time from the time of projection to reach a point at half of its maximum height while returning (in seconds) is
Answer: (B) $\left(1 + \dfrac{1}{\sqrt{2}}\right)t$
Rising to the maximum height $H$ takes time $t$, and falling from rest through $H$ also takes $t$:
$$H = \frac{1}{2}gt^2$$
On the way down, the body falls $\dfrac{H}{2}$ from the top to reach half the maximum height. If this takes $t'$:
$$\frac{H}{2} = \frac{1}{2}gt'^2 \;\Rightarrow\; t'^2 = \frac{t^2}{2} \;\Rightarrow\; t' = \frac{t}{\sqrt{2}}$$
Total time from projection:
$$t + \frac{t}{\sqrt{2}} = \left(1 + \frac{1}{\sqrt{2}}\right)t$$
Solution by Sreeraj P, M.Sc Physics