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

Answer 1

Answer: (B)

$2\sqrt{\frac{2\,H}{g}}$

Answer 2

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