Question

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

• 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 $[1+\frac{1}{\sqrt{2}}]t$

At $ma{x}^m$ height $v=o$,
$v=u-gt,\Rightarrow t=\frac{u}{g}=\sqrt{\frac{2H}{g}}-\left(1\right)$
Now situation is,
We want time to reach C.
Time from $AtoB=t$
Time from $BtoC\Rightarrow v_c^2=v_B^2+2g\frac{H}{2}$
$\frac{H}{2}=+\frac{1}{2}g{t}_1^2$ $v_C=\sqrt{gH}$
$t_1=\sqrt{\frac{H}{g}}\Rightarrow t_1=\frac{t}{\sqrt{2}}\left(from\right(1\left)\right)$
So, total time $=t+t_1=t+\frac{t}{\sqrt{2}}=t(1+\frac{1}{\sqrt{2}})$