Question #23db9

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Answer 1 · 2017-03-07T17:44:28

$7.70$ [m/s]

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

$p(t) = (v_x t, v_y t -1/2 g_m t^2)$

where $g_m$ is the free fall acceleration at the moon.

The velocity at time $t$ is given by

$v(t) = d/(dt)p(t) = (v_x, v_y-g_m t)$

after $t = 9$ [s] we have

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

$norm(v(9)) = sqrt(4^2+6.58^2) = 7.70$ [m/s]