Question

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. $\frac{2\sqrt{2\text{H sin}\theta }}{g}$
• D. $\frac{\sqrt{2\text{H sin}\theta }}{g}$

Answer 1

The correct option is B $2\sqrt{\frac{2H}{g}}$

We know that,
$\text{Maximum height},H=\frac{u^2{\text{sin}}^2\theta }{2g}$$\text{and Time of flight,}T=\frac{2u\text{sin}\theta }{g}\Rightarrow T^2=\frac{4u^2{\text{sin}}^2\theta }{g^2}$
$\therefore \frac{T^2}{H}=\frac{8}{g}\Rightarrow T=\sqrt{\frac{8H}{g}}=2\sqrt{\frac{2H}{g}}$