If the potential energy of a projectile at it | vedclass

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(C) Let the initial kinetic energy of the projectile be $E_0 = \frac{1}{2}mv_0^2$,where $v_0$ is the initial velocity.At maximum height,the potential

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let the initial kinetic energy of the projectile be $E_0 = \frac{1}{2}mv_0^2$,where $v_0$ is the initial velocity.At maximum height,the potential energy is $U = \frac{3}{4}E_0$.By the law of conservation of energy,the total energy remains constant. The kinetic energy at maximum height is $K = E_0 - U = E_0 - \frac{3}{4}E_0 = \frac{1}{4}E_0$.At maximum height,the vertical component of velocity is zero,so the kinetic energy is due to the horizontal component $v_x = v_0 \cos 	heta$.Thus,$K = \frac{1}{2}m(v_0 \cos 	heta)^2 = \frac{1}{2}mv_0^2 \cos^2 	heta = E_0 \cos^2 	heta$.Equating the two expressions for $K$: $E_0 \cos^2 	heta = \frac{1}{4}E_0$.$\cos^2 	heta = \frac{1}{4} \implies \cos 	heta = \frac{1}{2}$.Therefore,$	heta = 60^o$.

If the potential energy of a projectile at its maximum height is $3/4$ of its initial kinetic energy,then the angle of projection is ...... $^o$.

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