Question

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. $\sqrt{2}t$
• B. $1+\frac{1}{\sqrt{2}}t$
• C. $\frac{3t}{2}$
• D. $\frac{t}{\sqrt{2}}$

Answer 1

The correct option is B $1+\frac{1}{\sqrt{2}}t$

The ball is thrown vertically upwards then according to equation of motion
${\left(0\right)}^2-u^2=-2gh$ ...(i)
and $0=u-gt$ ...(ii)
From equations (i) and (ii),
$h=\frac{g{t}^2}{2}$
When the ball is falling downwards after reaching the maximum height
$s=u{t}^{\prime }+\frac{1}{2}g{\left(t^{\prime }\right)}^2$
$\frac{h}{2}=\left(0\right)t^{\prime }+\frac{1}{2}g{\left(t^{\prime }\right)}^2$
$\Rightarrow t^{\prime }=\frac{\sqrt{h}}{g}$
$t^{\prime }=\frac{t}{\sqrt{2}}$
Hence, the total time from the time of projection to reach a point at half of its maximum height while returning $=t+t^{\prime }$
$=t+\frac{t}{\sqrt{2}}$