Physics: Principles with Applications (7th Edition)

William should aim at an angle of $6.24^{\circ}$ above the horizontal.
We can use the equation for range $x$ to solve this question: $x = \frac{v_0^2~sin(2\theta)}{g}$ $sin(2\theta) = \frac{xg}{v_0^2}$ $sin(2\theta) = \frac{(27~m)(9.80~m/s^2)}{(35~m/s)^2}$ $sin(2\theta) = 0.216$ $\theta = \frac{arcsin(0.216)}{2}$ $\theta = 6.24^{\circ}$ Therefore, William should aim at an angle of $6.24^{\circ}$ above the horizontal.