Question

The maximum height reached by a projectile is 4 m. The horizontal range is 12 m. Velocity of projection in ms-1, is : ( g = acceleration due to gravity)

• A. $5\sqrt{\frac{g}{2}}$
• B. $5\frac{g}{\sqrt{2}}$
• C. $\frac{1}{3}\frac{g}{\sqrt{2}}$
• D. $\frac{1}{5}\sqrt{\frac{g}{2}}$

Answer 1

Answer: (A)

Given: Height of projectile H = 4m Horizontal range R = 12 m Height of projectile is $H=\frac{u^2{\sin }^2\theta }{2g}$ ?(i) Pange of projectile is $R=\frac{u^2\sin 2\theta }{g}$ ?(ii) Dividing equation (i) by (ii), also putting given value, we get $=\frac{H}{R}=\frac{u^2\sin \theta }{2g}/\frac{u^2\sin 2\theta }{g}$ $=\frac{u^2\sin \theta }{2g}\times \frac{g}{u^{2}2\sin \theta \cos \theta }$ $\frac{4}{12}=\frac{\tan \theta }{4}$ or $\tan \theta =\frac{4\times 4}{12}=\frac{4}{3}$ So, $\sin \theta =\frac{4}{5}$ Now, putting $H=4m$ and $\sin \theta =\frac{4}{5}s$ in Eq. (i) we get $H=\frac{u^2\times \frac{16}{5}}{2g}$ or $4=\frac{u^2\times 16/25}{2g}$ $u^2=4\times 2g\times \frac{25}{16}$ Hence, $u=5\sqrt{\frac{g}{2}}m/s$