SOLUTION: When an object is dropped from a height of 20.0 m above the surface of Planet Z, it will fall 5.00 m during the 2nd second of the fall. What is the acceleration near the surface o

Let  "a" be the unknown acceleration under the question, in m/s^2.


Then your equation is

 -  = 5.   (1)


The term   is the distance that freely falling body passes in two seconds.  (Remember the formula S =  from Physics ?)

The term   is the distance that freely falling body passes in the first second.

The difference is what you are given.


From the equation (1) you have 

4a - a = 2*5,   or

3a = 10,   which gives you  a =  =  m/s^2.



The given value of the height 20 m  IS NOT RELEVANT  to the solution.


We can use it only to check whether our body will be still falling during 2 seconds:

 =  =  = 7 m.


Yes, it still be falling.


So, we solved the problem correctly.


Answer.  The acceleration near the surface of the planet is  =  m/s^2.

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







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 v0 = a.
The 2nd second lasts for 1 second, so we substitute 1 for the time t: 















So the acceleration of gravity for Planet Z is -3.33

Edwin