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) $\left(1 + \dfrac{1}{\sqrt{2}}\right)t$
Concept. Rising to the top takes the same time as falling back from rest through the same height: $H = \dfrac12gt^2$.
Step 1. Fall to half height. Falling $\dfrac{H}{2}$ from rest takes $t'$:
$$\frac{H}{2} = \frac12gt'^2 \;\Rightarrow\; t' = \frac{t}{\sqrt2}$$
Step 2. Total time.
$$t + \frac{t}{\sqrt2} = \left(1 + \frac{1}{\sqrt2}\right)t$$
Answer. $\left(1 + \dfrac{1}{\sqrt2}\right)t$ (Option B).
Solution by Sreeraj P, M.Sc Physics