Answer
a) 28.9mph
b) 13mph
Work Step by Step
a) We proceed to use conservation of momentum
$p_{x}$= ((6500lb)/g)$v_{0}$ =(9452lb/g)$v_{w}$sin(39$^{\circ})$
$p_{y}$= ((3402lb)/g)(50mph) =(9452lb/g)$v_{w}$sin(39$^{\circ})$
Solving for $v_{0}$ gives 28.9mph
b)We proceed to use energy conservation
$U_{1}+K_{1}+W_{others}= U_{2}+K_{2}$
$\frac{1}{2}mv_{1}^{2} - \mu_{k}mgd = \frac{1}{2}mv_{2}^{2}$
$v_{2}=\sqrt(v_{1}^{2} -2\mu_{k}gd)$
$v_{1}$=37.1ft/s, g=32.2ft/s$^{2}$, d=35ft, we get $v_{2}$=19.1ft/s = 13mph
