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(D) The initial kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.At the maximum height,the vertical component of velocity is zero,so the velocity

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$\cos^{-1} (\sqrt{3/10})$

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(D) The initial kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.At the maximum height,the vertical component of velocity is zero,so the velocity is only the horizontal component $v_x = v \cos 	heta$.The kinetic energy at the maximum height $K'$ is given by $K' = \frac{1}{2}m(v \cos 	heta)^2 = K \cos^2 	heta$.Given $K = 100 \, J$ and $K' = 30 \, J$,we have:$30 = 100 \cos^2 	heta$$\cos^2 	heta = \frac{30}{100} = \frac{3}{10}$$\cos 	heta = \sqrt{\frac{3}{10}}$$	heta = \cos^{-1} \left( \sqrt{\frac{3}{10}} ight)$.

If the initial kinetic energy of a projectile is $100 \, J$ and the kinetic energy at the maximum height is $30 \, J$,what is the angle of projection?

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