Answer
$(a)\space 3.7\space m/s$
$(b)\space 68.7^{\circ}$
Work Step by Step
Please see the image first.
We know, at the maximum height, the vertical component of velocity becomes zero. So let's apply equation $S=\frac{(u+V)}{2}t$ into vertical direction.
$\uparrow S=\frac{(u+V)}{2}t$
Let's plug known values into this equation.
$0.62\space m=\frac{(Vsin\theta+0)}{2}t\space =\gt\space 1.24=Vtsin\theta-(1)$
Let's apply equation $S=ut$ in the horizontal direction.
$\rightarrow S=ut$
Let's plug known values into this equation.
$0.483=Vcos\theta\space t-(2)$
$(1)\div(2)$
$\frac{1.24}{0.483}=tan\theta\space =\gt \theta=68.7^{\circ}$
Let's apply equation $V^{2}=u^{2}+2aS$ in vertical direction.
$\uparrow V^{2}=u^{2}+2aS$
Let's plug known values into this equation.
$0=(Vsin68.7^{\circ})^{2}+2\times (-9.8\space m/s^{2})\times0.62\space s$
$12.15\space m^{2}/s^{2}= V^{2}\times 0.87$
$\sqrt {13.96}\space m/s=V$
$V=3.7\space m/s$
