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?

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Answer 1 · 2015-07-01T16:11:12

I fond: $11.5m/s$ downwards

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.

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^2s^2, what is the probe's velocity when it reaches the ground?