Me

s=1/2at^2
the formula shown represents the distance s that a free-falling object will fall near a planet or the moon in a given time t. in the formula, a represents the acceleration due to gravity.

1) solve the formula for a.

2) solve the formula for t.

3) a free-falling object near the moon drops 20.5 meters in 5 seconds. what is the acceleration a for the moon?

4) using this value for a, determine how many seconds it will take for an object to fall 42.5 meters.

Just use algebra to rearrange the terms so for s=1/2at^2 and solving for a we get 2s=at^2 so 2s/t^2=a
solving s=1/2at^2 for t gives 2s=at^2 which leads to 2s/a=t^2 so t=sqrt(2s/a)

filling in the numbers for 3 since a = 2s/t^2 we get a= (2(20.5m))/5s^2 =1.64m/s^2 the correct units in the answer indicate a possible correct answer since the units of acceleration are m/s^2.

4. using the value for a from question 3 and using 42.5 meters for s and solving for t we get sqrt((2s)/a) = sqrt((2*42.5m)/1.64m/s^2)= approximately 7.2s you can see the m's cancel and you are left s^2 as the units and the sqrt(s^2) = s so again the correct units indicate that 7.2s is probably correct. (These answers ARE correct though). checking the units gives a kinda quick assurance you are on the right path.