Question

A body is thrown vertically up to reach its maximum height in 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. $\sqrt{2t}$
• B. $(1+\frac{1}{\sqrt{2}})t$
• C. $\frac{3t}{2}$
• D. $\frac{t}{\sqrt{2}}$

Answer 1

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

Time taken to reach the maximum height $=$ Time of ascent $=t$

Therefore,

As $h\propto t^2⟹\frac{h_1}{h_2}=(\frac{t_1}{t_2{)}^2}⟹\frac{h}{h/2}=(\frac{t}{t_2}{)}^2$

$⟹t_2=\frac{t}{\sqrt{2}}$

Time taken toreach half distance of its maximum height $=\frac{1}{\sqrt{2}}$

So, Total time $=t+\frac{t}{\sqrt{2}}=(1+\frac{1}{\sqrt{2}})t$