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

The Correct Option is B

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 $H =\frac{1}{2 g}\left(\frac{T g}{2}\right)^{2}$ $H=\frac{T^{2} g^{2}}{8 g}$ $T^{2} =\frac{8 H}{g}$ $T=2 \sqrt{\left(\frac{2 H}{g}\right)}$