# A surface probe on the planet Mercury falls 17.6m downward from a ledge. If free-fall acceleration near Mercury is -3.70 m/s^2, what is the probe's velocity when it reaches the ground?

Jul 1, 2015

I fond: $11.5 \frac{m}{s}$ downwards

#### Explanation:

We can first find the time it takes to hit the ground (assuming starting with ${v}_{i} = 0$) using:
${y}_{f} - {y}_{i} = {v}_{i} t + \frac{1}{2} \cdot a {t}^{2}$ so:

$0 - 17.6 = 0 - \frac{3.70}{2} {t}^{2}$
$t = 3.1$sec.

Now we use: ${v}_{f} = {v}_{i} + a t$ inserting the time found before:
${v}_{f} = 0 - 3.70 \cdot 3.10 = - 11.5 \frac{m}{s}$
The minus sign telling us that it is directed dawnwards.