# Question #23db9

Mar 7, 2017

$7.70$ [m/s]

#### Explanation:

Taking the astronaut position as the referential origin, the ball trajectory is given by

$p \left(t\right) = \left({v}_{x} t , {v}_{y} t - \frac{1}{2} {g}_{m} {t}^{2}\right)$

where ${g}_{m}$ is the free fall acceleration at the moon.

The velocity at time $t$ is given by

$v \left(t\right) = \frac{d}{\mathrm{dt}} p \left(t\right) = \left({v}_{x} , {v}_{y} - {g}_{m} t\right)$

after $t = 9$[s] we have

$v \left(9\right) = \left(4 , 8 - 1.62 \cdot 9\right) = \left(4 , - 6.58\right)$ and the speed

$\left\lVert v \left(9\right) \right\rVert = \sqrt{{4}^{2} + {6.58}^{2}} = 7.70$ [m/s]