Answer
(a) The arrow will miss the target by 13.3 meters.
(b) $\theta = 22.2^{\circ}$
Work Step by Step
(a) $t = \frac{x}{v_x} = \frac{38.0 ~m}{23.1 ~m/s} = 1.65 ~s$
We can use the time t to find how far the arrow will drop before it reaches the target.
$y = \frac{1}{2}at^2 = \frac{1}{2}(9.80 ~m/s^2)(1.65 ~s)^2 = 13.3 ~m$
(b) $x = \frac{v^2 ~sin(2\theta)}{g}$
$sin(2\theta) = \frac{gx}{v^2} = \frac{(9.80 ~m/s^2)(38.0 ~m)}{(23.1 ~m/s)^2} = 0.698$
$2\theta = sin^{-1}(0.698) = 44.3^{\circ}$
$\theta = \frac{44.3^{\circ}}{2} = 22.2^{\circ}$
