Question 1087593
.
<pre>
Let  "a" be the unknown acceleration under the question, in m/s^2.


Then your equation is

{{{(a*2^2)/2}}} - {{{(a*1^2)/2}}} = 5.   (1)


The term  {{{(a*2^2)/2}}} is the distance that freely falling body passes in two seconds.  (Remember the formula S = {{{(at^2)/2}}} from Physics ?)

The term  {{{(a*1^2)/2}}} 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 = {{{10/3}}} = {{{3}}}{{{1/3}}} m/s^2.



The given value of the height 20 m  <U>IS NOT RELEVANT</U>  to the solution.


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

{{{(a*t^2)/2}}} = {{{(((10/3))*2^2)/2}}} = {{{20/3}}} = 7 m.


Yes, it still be falling.


So, we solved the problem correctly.


<U>Answer</U>.  The acceleration near the surface of the planet is {{{10/3}}} = {{{3}}}{{{1/3}}} m/s^2.
</pre>

Solved.