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?

1 Answer
Jul 1, 2015

Answer:

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.