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=4\,m $ 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 $