Q. A stone is 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) $\frac{\sqrt{2H\sin\theta}}{g}$
Explanation
(b) $H = \frac{u^{2}\sin^{2}\theta}{2g}\ and\ T = \frac{2u\sin\theta}{g}\ \Rightarrow T^{2} = \frac{4\ u^{2}\sin^{2}\theta}{g^{2}}$
$\therefore\frac{T^{2}}{H} = \frac{8}{g} \Rightarrow T = \sqrt{\frac{8H}{g}} = 2\sqrt{\frac{2H}{g}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.