Question

A body is projected at an angle $\theta$ so that its range is maximum. If $T$ is the time of flight then the value of maximum range is:

[acceleration due to gravity$=$g]

• A. $\frac{g^{2}T}{2}$
• B. $\frac{gT}{2}$
• C. $\frac{g{T}^2}{2}$
• D. $\frac{g^2{T}^2}{2}$

Answer 1

The correct option is C $\frac{g{T}^2}{2}$

Range is maximum when $\theta ={45}^o$

Time of flight $T=\frac{2usin\theta }{g}$

$\therefore$ $T=\frac{2usin{45}^o}{g}=\frac{\sqrt{2}u}{g}$ $⟹$ $u=\frac{gT}{\sqrt{2}}$

Maximum range $R_{max}=\frac{u^2}{g}=\frac{\frac{g^2{T}^2}{2}}{g}=\frac{g{T}^2}{2}$