Answer
2128 N at 42.6 degrees with the horizontal
Work Step by Step
1. KE(initial) = PE(final) (Conservation of Energy)
2. $\frac{1}{2}$m$v^{2}$ = mgy (KE = PE)
3. V$_{final}$ = $\sqrt(2g\Delta y)$ = $\sqrt (2*9.8*31.5)$ = 24.85 m/s (Solve for Velocity)
4. F$_{x}$ = $\frac{\Delta p_{x}}{\Delta t}$ = $\frac{m\Delta v_{x}}{\Delta t}$ = $\frac{.145(27)}{2.5\times10^{-3}}$ = 1566 N (Solve for Force in x direction)
5. F$_{y}$ = $\frac{\Delta p_{y}}{\Delta t}$ = $\frac{m\Delta v_{y}}{\Delta t}$ = $\frac{.145(24.85)}{2.5\times10^{-3}}$ = 1441 N (Solve for Force in y direction)
6. F = $\sqrt (F_{x}^{2} + F_{y}^{2})$ = $\sqrt (1566^{2} + 1441^{2})$ = 2128 N (Use Pythagorean Theorem to find the total force)
7. $\theta$ = $tan^{-1}(\frac{1441}{1566})$ = 42.6$^{\circ}$ with the horizontal (Magnitude)
