Answer
$\theta = 14.3^{\circ}$
Work Step by Step
When $x = 0$, the vertical component of velocity is $v_y = 5.0~m/s$
Therefore, $v_{y0} = 5.0~m/s$
When $x = 10~m$, $v_y = 0$ when the ball hits the wall.
We can find the time when $v_y = 0$:
$v_{yf} = v_{y0}+at$
$t = \frac{v_{yf} - v_{y0}}{a}$
$t = \frac{0 - 5.0~m/s}{-9.8~m/s^2}$
$t = 0.51~s$
We can find $v_x$:
$x = v_x~t$
$v_x = \frac{x}{t}$
$v_x = \frac{10~m}{0.51~s}$
$v_x = 19.6~m/s$
We can find the launch angle:
$tan~\theta = \frac{v_y}{v_x}$
$tan~\theta = \frac{5.0~m/s}{19.6~m/s}$
$\theta = tan^{-1}~\frac{5.0~m/s}{19.6~m/s}$
$\theta = 14.3^{\circ}$
