Answer
$v_2=\sqrt{v_1^2+2gh}$
$ \theta_2=sin^{-1}(\frac{v_1 sin\theta_1}{\sqrt{v_1^2+2gh}})$
Work Step by Step
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
$mv_1sin\theta_1=mv_2sin\theta_2$
$\implies sin\theta_2=\frac{v_1sin\theta_1}{\sqrt{v_1^2+2gh}}$
$\implies \theta_2=sin^{-1}(\frac{v_1 sin\theta_1}{\sqrt{v_1^2+2gh}})$
