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Question

(B) Let the initial velocity be $v_1$. The initial kinetic energy is $K = \frac{1}{2} m v_1^2$.In projectile motion,the horizontal component of veloci

Vedclass · Accepted Answer

$K\left( 1 - \frac{\cos^2 	heta_1}{\cos^2 	heta_2} ight)$

Vedclass · Answer

(B) Let the initial velocity be $v_1$. The initial kinetic energy is $K = \frac{1}{2} m v_1^2$.In projectile motion,the horizontal component of velocity remains constant throughout the flight.Therefore,$v_1 \cos 	heta_1 = v_2 \cos 	heta_2$,where $v_2$ is the velocity at angle $	heta_2$.From this,$v_2 = \frac{v_1 \cos 	heta_1}{\cos 	heta_2}$.The kinetic energy at angle $	heta_2$ is $K' = \frac{1}{2} m v_2^2 = \frac{1}{2} m \left( \frac{v_1 \cos 	heta_1}{\cos 	heta_2} ight)^2$.$K' = \left( \frac{1}{2} m v_1^2 ight) \frac{\cos^2 	heta_1}{\cos^2 	heta_2} = K \frac{\cos^2 	heta_1}{\cos^2 	heta_2}$.By the law of conservation of mechanical energy,the gain in potential energy $(PE)$ is equal to the loss in kinetic energy.$PE = K - K' = K - K \frac{\cos^2 	heta_1}{\cos^2 	heta_2} = K \left( 1 - \frac{\cos^2 	heta_1}{\cos^2 	heta_2} ight)$.

$A$ body is projected from the ground at an angle $\theta_1$ with the horizontal with kinetic energy $K$. What is the potential energy of the body when the velocity makes an angle $\theta_2$ with the horizontal?

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