Answer
a) 37.7 degrees
b) $-.657\ m/s$
Work Step by Step
a) This is a matter of kinematics. We first use the equation for change in y:
$\Delta y = v_{0y}t + \frac{1}{2}gt^2$
$0 = v_{0y}t + \frac{1}{2}gt^2$
$-v_{0y}t = \frac{1}{2}gt^2$
$v_{0y} = -\frac{1}{2}gt$
$t = \frac{-2v_0 y}{g}$
We also find:
$\Delta x = v_{0x}t$
Thus, it follows:
$15.2 = v_{0x}\frac{-2v_0 y}{g}$
$74.56=v_{0x}v_{0y}$
$74.56=v_{0}^2sin\theta cos\theta$
$.51775=sin\theta cos\theta$
$\theta=37.7^{\circ}$
b) We use conservation of momentum:
$0 = (12)(4.5)cos(37.7^{\circ})+65v\\ v=-.657\ m/s$
