# Molly kicks a soccer ball into the air with an initial velocity of 15 m/s. It lands 20 meters from where she kicked it. At what angle did Molly launch the ball?

Feb 7, 2016

$\theta = \frac{1}{2} {\sin}^{-} 1 \left(\frac{20}{225}\right) \text{radians}$

#### Explanation: The x and y components of initial velocity ${v}_{o} = 15 \frac{m}{s}$ are
1. v_x = v_o cos theta;  and
2. ${v}_{y} = {v}_{o} \sin \theta - \text{gt}$
3. from 1) the distance in x is $x \left(t\right) = {v}_{o} t \cos \theta$
a) Total distance in x, Range $R = 20 = x \left({t}_{d}\right) = {v}_{o} {t}_{\mathrm{dc}} o s \theta$
b) Where ${t}_{d}$ is the total distance required to travel R = 20 m
4. The displacement in y is $y \left(t\right) = {v}_{o} t \sin \theta - \frac{1}{2} {\text{gt}}^{2}$
a) at time t = t_d; y(t_d) = 0
b) setting y=0 and solving for time, ${t}_{d} = 2 {v}_{o} \sin \frac{\theta}{g}$
5. Insert 4.a) into 3.a) we get, $R = 2 {v}_{o}^{2} \frac{\cos \theta \sin \theta}{g}$
a) 5. above can also be written as: $R = {v}_{o}^{2} / g \sin 2 \theta$

Now we know, R = 20m; v_o = 15m/s solve for $\theta$
$\theta = \frac{1}{2} {\sin}^{-} 1 \left(\frac{20}{225}\right) \text{radians}$