The angle of projection of a projectile for w | vedclass

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Question

(C) Let the initial velocity of the projectile be $u$ and the angle of projection be $	heta$. The initial kinetic energy is $K_i = \frac{1}{2}mu^2$.A

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) Let the initial velocity of the projectile be $u$ and the angle of projection be $	heta$. The initial kinetic energy is $K_i = \frac{1}{2}mu^2$.At the maximum height, the vertical component of velocity becomes zero, and the velocity of the projectile is only the horizontal component, $v_x = u \cos 	heta$.The kinetic energy at the maximum height is $K_f = \frac{1}{2}m(u \cos 	heta)^2 = \frac{1}{2}mu^2 \cos^2 	heta$.According to the problem, $K_f = \frac{1}{2}K_i$.Substituting the expressions, we get $\frac{1}{2}mu^2 \cos^2 	heta = \frac{1}{2} (\frac{1}{2}mu^2)$.This simplifies to $\cos^2 	heta = \frac{1}{2}$, which means $\cos 	heta = \frac{1}{\sqrt{2}}$.Therefore, $	heta = 45^{\circ}$.

The angle of projection of a projectile for which its initial kinetic energy becomes half at its maximum height is (in $^{\circ}$)

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