Q123 / 160 · TS EAMCET 2008 · Physics
A body thrown vertically up to reach its maximum height in $t$ second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is
- A$\sqrt{2} t$
- Correct answerB$\left(1+\frac{1}{\sqrt{2}}\right) t$
- C$\frac{3 t}{2}$
- D$\frac{t}{\sqrt{2}}$
Solution
The ball is thrown vertically upwards then according to equation of motion
$$
(0)^2-u^2=-2 g h
$$
and
$$
0=u-g t
$$
From Eqs. (i) and (ii),
$$
h=\frac{g t^2}{2}
$$
When the ball is falling downwards after reaching the maximum height
$$
\begin{aligned}
s & =u t^{\prime}+\frac{1}{2} g\left(t^{\prime}\right)^2 \\
\frac{h}{2} & =(0) t^{\prime}+\frac{1}{2} g\left(t^{\prime}\right)^2 \\
t^{\prime} & =\sqrt{\frac{h}{g}} \\
t^{\prime} & =\frac{t}{\sqrt{2}}
\end{aligned}
$$
Hence, the total time from the time of projection to reach a point at half of its maximum height while returning $=t+t^{\prime}$
$$
=t+\frac{t}{\sqrt{2}}
$$
Hence, the correct answer is option: 'B'.