Question #45954

Mar 5, 2017

$\theta \approx {42}^{\circ}$

Explanation:

Note that it takes $\frac{30}{2} = 15$ seconds for the vertical component of the velocity to become $0$.

Denoting the initial vertical component of the velocity by $v$ and using $g = 9.8 m {s}^{- 2}$, we have:

$v = 9.8 \cdot 15 = 147$

So the launch angle $\theta$ satisfies:

$\sin \left(\theta\right) = \frac{147}{220}$

So $\theta = {\sin}^{- 1} \left(\frac{147}{220}\right) \approx {42}^{\circ}$