Question

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?

Answer 1

I fond: `11.5m/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_it+1/2*at^2` so:

`0-17.6=0-3.70/2t^2`
`t=3.1`sec.

Now we use: `v_f=v_i+at` inserting the time found before:
`v_f=0-3.70*3.10=-11.5 m/s`
The minus sign telling us that it is directed dawnwards.