A projectile is fired with KE of 1,kJ. If the | vedclass

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(B) For maximum range,the angle of projection $	heta$ must be $45^o$.Initial kinetic energy $KE_i = \frac{1}{2}mv^2 = 1000\,J$.At the highest point,t

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$500$

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(B) For maximum range,the angle of projection $	heta$ must be $45^o$.Initial kinetic energy $KE_i = \frac{1}{2}mv^2 = 1000\,J$.At the highest point,the vertical component of velocity is zero,so the velocity is only the horizontal component $v_x = v \cos(45^o) = \frac{v}{\sqrt{2}}$.The kinetic energy at the highest point is $KE_f = \frac{1}{2}m(v_x)^2 = \frac{1}{2}m(\frac{v}{\sqrt{2}})^2 = \frac{1}{2}m(\frac{v^2}{2}) = \frac{1}{2} KE_i$.Therefore,$KE_f = \frac{1000}{2} = 500\,J$.

$A$ projectile is fired with $KE$ of $1\,kJ$. If the range is maximum,.......... $J$ is its $KE$ at the highest point.

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