## Engineering Mechanics: Statics & Dynamics (14th Edition)

$v_2=\sqrt{v_1^2+2gh}$ $\theta_2=sin^{-1}(\frac{v_1 sin\theta_1}{\sqrt{v_1^2+2gh}})$
The required speed and angle can be determined as follows: According to the law of conservation of energy $\frac{1}{2}mv_1^2+mgh=\frac{1}{2}mv_2^2+0$ This simplifies to: $v_2=\sqrt{v_1^2+2gh}$ Now, according to the principle of impulse and momentum $mv_1sin\theta_1=mv_2sin\theta_2$ $\implies sin\theta_2=\frac{v_1sin\theta_1}{\sqrt{v_1^2+2gh}}$ $\implies \theta_2=sin^{-1}(\frac{v_1 sin\theta_1}{\sqrt{v_1^2+2gh}})$