Question

A stone thrown at an angle $\theta$ to the horizontal reaches a maximum height $H$ . Then the time of flight of stone will be

• A. $\sqrt{\frac{2H}{g}}$
• B. $2\sqrt{\frac{2H}{g}}$
• C. $\frac{2\sqrt{2H\sin \theta }}{g}$
• D. $\frac{\sqrt{2H\sin \theta }}{g}$

Answer 1

Answer: (B)

Maximum height
$H=\frac{u^2{\sin }^2\theta }{2g}$ ...(1)
Time of flight
$T=\frac{2u\sin \theta }{g}$
$\Rightarrow u\sin \theta =\frac{Tg}{2}$ ...(2)
From equations (1) and (2), we get
$H=\frac{1}{2g}{\left(\frac{Tg}{2}\right)}^2$
$H=\frac{T^2{g}^2}{8g}$
$T^2=\frac{8H}{g}$
$T=2\sqrt{\left(\frac{2H}{g}\right)}$