Question I

I) A Moon rock is thrown upward with velocity 7 m/s. After 7 s, it has a downward velocity of 4 m/s. What is the acceleration due to gravity on the Moon (10 pts)? How high above the starting point did the rock go before it began to fall (10 pts)?

Solution:
We know that
v_f = v₀ - g_moon t
Plugging in we find that
$g_{moon} = \frac{v_{0}-v_{f}}{t}$
$g_{moon}= \frac{7-(-4)}{7} ~=~ -1.6 m/s^{2}$
So, if we want the distance $\Delta y$ that the rock rises, we use the relation
$v_{f}^{2}=v_{0}^{2}-2a\Delta y$
In our case, v_f = 0, so we have
$\Delta y = \frac{v_{0}^{2}}{2g_{moon}}$
Plugging in the number here yields
$\Delta y = 16 m$.