Question 1087593
<pre><b>
Let's look at what happens during the very first second.  
The initial velocity is 0, since it is 'dropped', rather
than thrown down.

{{{v[F] = v[0]+at}}}

{{{v[F]=0+a(1)}}}

{{{v[F]=a]}}}

Now we look at what happens during the 2nd second.

The final velocity during the first second, which is 'a', will be
the initial velocity during the 2nd second.  We are told that the
distance it falls during the 2nd second is x = -5.00 m (negative
because it's falling downward.  So we substitute that and v<sub>0</sub> = a.
The 2nd second lasts for 1 second, so we substitute 1 for the time t: 

{{{x=v[0]t+expr(1/2)at^2}}}

{{{-5.00=(a)(1)+expr(1/2)a(1)^2}}}

{{{-5=a+expr(1/2)a}}}

{{{-5=expr(3/2)a}}}

{{{-10=3a}}}

{{{-10/3=a}}}

{{{-3.33=a}}}

So the acceleration of gravity for Planet Z is -3.33{{{m/s^2}}}

Edwin</pre>