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

(1) $\sqrt{\frac{2H}{g}}$

(2) $2\text{\hspace{0.17em}}\sqrt{\frac{2H}{g}}$

(3) $\frac{2\sqrt{2H\text{\hspace{0.17em}}\mathrm{sin}\theta }}{g}$

(4) $\frac{\sqrt{2H\text{\hspace{0.17em}}\mathrm{sin}\theta }}{g}$

Concept Videos :-

#11-Projectile-Motion-1
#12-Projectile-Examples
#13-Projectile-Motion-2
#14-Projectile-Examples-1
#15-Projectile-Examples-2
#16-Solved-Examples-3
#17-Projectile-Examples-4
#18-Maximum-Height
#20-Derivation-of-Equation-of-Trajectory
#21-Solved-Examples-11

Concept Questions :-

Projectile motion

(2) $H=\frac{{u}^{2}{\mathrm{sin}}^{2}\theta }{2g}$ and $T=\frac{2u\mathrm{sin}\theta }{g}$${T}^{2}=\frac{4{u}^{2}{\mathrm{sin}}^{2}\theta }{{g}^{2}}$

$\frac{{T}^{2}}{H}=\frac{8}{g}$$T=\sqrt{\frac{8H}{g}}=2\sqrt{\frac{2H}{g}}$

Difficulty Level:

• 36%
• 49%
• 11%
• 6%