Q. A stone is thrown at an angle $\theta$ to the horizontal reaches a maximum heights 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) $Maximum\ height\ H = \frac{u^{2}\sin^{2}\theta}{2g}$
$and\ times\ of\ flight\ T = \frac{2usin\theta}{g}$
$or\ \ T^{2} = \frac{4u^{2}\sin^{2}\theta}{g^{2}}$
$\therefore\ \frac{T^{2}}{H} = \frac{8}{g}$
$or\ T = \sqrt{\frac{8H}{g}} = 2\sqrt{\frac{2H}{g}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.