Question

The speed of a projective when it is at its greatest height is $\sqrt{\frac{2}{5}}$ times its speed at half the maximum height, What is its angle of projection?

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

Maximum height $=\frac{u^{2}si{n}^2\theta }{2g}$
Half the maximum height $=\frac{u^{2}si{n}^2\theta }{2g}$
Horizontal velocity at half the maximum height $=ucos\theta$
Vertical velocity at half the maximum height $=\sqrt{u^{2}si{n}^2\theta -2g\left(\frac{usin\theta }{2g}\right)}=\frac{usin\theta }{\sqrt{2}}$
Velocity at half the maximum height $=\sqrt{u^{2}si{n}^2\theta \frac{u^{2}si{n}^2\theta }{2}}ucos\theta =\sqrt{\frac{2}{5}}\sqrt{u^{2}si{n}^2\theta \frac{u^{2}si{n}^2\theta }{2}}\Rightarrow u^{2}co{s}^2\theta =\frac{2}{5}{u}^2(1-si{n}^2\theta +\frac{si{n}^2\theta }{2})\Rightarrow 5(1-si{n}^2\theta =2-si{n}^2\theta )\Rightarrow 3=4si{n}^2\theta \Rightarrow sin\theta =\frac{\sqrt{3}}{2}\Rightarrow \theta =\frac{\pi }{3}$.