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(C) Let the initial velocity be $u$. The velocity at the highest point is $v_H = u \cos \alpha$.The maximum height is $H = \frac{u^2 \sin^2 \alpha}{2g

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let the initial velocity be $u$. The velocity at the highest point is $v_H = u \cos \alpha$.The maximum height is $H = \frac{u^2 \sin^2 \alpha}{2g}$.At height $h = \frac{H}{2} = \frac{u^2 \sin^2 \alpha}{4g}$,the horizontal component of velocity is $v_x = u \cos \alpha$ and the vertical component is $v_y = \sqrt{u^2 \sin^2 \alpha - 2gh}$.Substituting $h$,we get $v_y = \sqrt{u^2 \sin^2 \alpha - 2g(\frac{u^2 \sin^2 \alpha}{4g})} = \sqrt{u^2 \sin^2 \alpha - \frac{u^2 \sin^2 \alpha}{2}} = \frac{u \sin \alpha}{\sqrt{2}}$.The resultant velocity at height $h$ is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 \cos^2 \alpha + \frac{u^2 \sin^2 \alpha}{2}}$.Given $v_H = \sqrt{\frac{2}{5}} v$,so $v_H^2 = \frac{2}{5} v^2$.$u^2 \cos^2 \alpha = \frac{2}{5} (u^2 \cos^2 \alpha + \frac{u^2 \sin^2 \alpha}{2})$.$u^2 \cos^2 \alpha = \frac{2}{5} u^2 \cos^2 \alpha + \frac{1}{5} u^2 \sin^2 \alpha$.$\frac{3}{5} \cos^2 \alpha = \frac{1}{5} \sin^2 \alpha$.$	an^2 \alpha = 3 \implies 	an \alpha = \sqrt{3}$.Therefore,$\alpha = 60^{\circ}$.

$A$ projectile is thrown with some initial velocity at an angle $\alpha$ to the horizontal. Its velocity when it is at the highest point is $(2/5)^{1/2}$ times the velocity when it is at height half of the maximum height. Find the angle of projection $\alpha$ with the horizontal. (in $^{\circ}$)

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