Oct 3rd 2017, 06:34 PM #2 Senior Member     Join Date: Aug 2008 Posts: 113 Have you performed the conversions as directed? (a) use the equations $\Delta x = v_0 \cos{\theta} \cdot t$ to determine the time to reach the center field wall, then the equation $\Delta y = v_0 \sin{\theta} \cdot t - \dfrac{1}{2}gt^2$ to determine if the ball clears the wall. (b) maximum height occurs when the y-component of velocity equals zero ... $0 = v_0 \sin{\theta} - gt$, solve for $t$, then use the $\Delta y$ equation to determine the max height. (c) set $\Delta y = -4 \, ft$ and solve for $t$ (remember to convert feet to meters first) (d) get some graph paper and do as instructed ... you can eliminate the parameter of time to get $y$ as a function of $x$.