# A water balloon with a mass of 500 g is launched across a waterpark and lands 150 m away 5 seconds later. What is the initial velocity in the x and y directions, launch angle, and max height?

Apr 24, 2018

The launch angle was approximately ${39.2}^{\circ}$. The vertical launch velocity was 24.5 m/s. The horizontal launch velocity was 30 m/s. The maximum height was 30.625 m.

#### Explanation:

Let

$\theta$ = the launch angle

$v$ = the launch velocity

$g$ = the acceleration of gravity = 9.8 $\frac{m}{s} ^ 2$

Assume no air resistance and that the water balloon was launched at ground level.

Horizontal velocity = $v \cos \theta$

We know that the balloon traveled 150 meters horizontally in 5 seconds so

$v \cos \theta = \frac{150}{5} = 30$ m/s

The horizontal distance, $x$, as a function of time, $t$, is

$x = 30 t$

which implies that

$t = \frac{x}{30}$.

The vertical height, $y$, as a function of time, $t$, is

$y = v \left(\sin \theta\right) t - \frac{g}{2} {t}^{2}$

Use the last two equations to eliminate $t$.

$y = v \left(\sin \theta\right) \frac{x}{30} - \frac{g}{2} {x}^{2} / 900$

The balloon lands when $x = 150$, so $y = 0$ when $x = 150$.

$0 = v \left(\sin \theta\right) \frac{150}{30} - \frac{g}{2} \left({150}^{2} / 900\right)$

$v \left(\sin \theta\right) = \frac{g}{2} \left(\frac{30}{150}\right) \left({150}^{2} / 900\right) = \frac{9.8 \left(150\right)}{2 \left(30\right)} = \frac{9.8 \left(5\right)}{2} = 24.5$ m/s

$v \left(\sin \theta\right) = 24.5$ m/s is the vertical launch velocity.

Notice that

$\tan \theta = \frac{v \sin \theta}{v \cos \theta} = \frac{24.5}{30}$

so

$\theta = {\tan}^{-} 1 \left(\frac{24.5}{30}\right) \approx {39.2}^{\circ}$

The max height will be halfway through the flight or when $t = \frac{5}{2}$.

Remember that $y = v \left(\sin \theta\right) t - \frac{g}{2} {t}^{2}$, so the max height is

$y = 24.5 \left(\frac{5}{2}\right) - \frac{9.8}{2} {\left(\frac{5}{2}\right)}^{2} = 30.625$ m.