Question

A ball falling from rest is located 45m below its starting point 3.0s later. Assuming that its acceleration is uniform, what is its value?

I know that the appropriate equation is
∆d = vi∆t + 1/2a∆t^2, but I do not understand why.

Answer 1

The acceleration is `10 (m /s^2)`

Explanation:

Let the acceleration be `a`
Then we can use the equation `s=ut+1/2at^2`
`u=0` initial velocity, starting frm rest
`s=45 m` distance
`t=3 s` time
So `45=0+1/2a*9`
Aussuming that the downward direction is positive
`9/2a=45`
So `a=2*45/9=2*5=10 (m /s^2)`
This is `g`