Essential University Physics: Volume 1 (4th Edition)

We first find the change in time: $55=28cos(9.5)t \\ t = 1.99 \ s$ We now find the change in y: $\Delta y = -28sin(9.5)(1.99)-\frac{1}{2}g(1.99)^2=-28.65 \ m$ Thus, we find the angle they will be at: $\theta=tan^{-1}(\frac{28.65}{55})=27^{\circ}$ This means that the slope needs to be 30 degrees below the horizontal.