#### Answer

William should aim at an angle of $6.24^{\circ}$ above the horizontal.

#### Work Step by Step

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.