At the highest point of the path of a project | vedclass

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(C) For a projectile launched with an initial velocity $u$ at an angle $	heta$,the velocity at any point is given by $v = \sqrt{v_x^2 + v_y^2}$.At th

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Kinetic energy is minimum

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(C) For a projectile launched with an initial velocity $u$ at an angle $	heta$,the velocity at any point is given by $v = \sqrt{v_x^2 + v_y^2}$.At the highest point of the trajectory,the vertical component of velocity $v_y$ becomes $0$,while the horizontal component $v_x = u \cos 	heta$ remains constant.Since the kinetic energy is given by $K = \frac{1}{2}mv^2$,it is proportional to the square of the velocity.At the highest point,the velocity is minimum $(v = v_x = u \cos 	heta)$,therefore the kinetic energy is minimum.Conversely,the potential energy $U = mgh$ is maximum at the highest point because the height $h$ is maximum.

At the highest point of the path of a projectile,its

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