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.

Oct 21, 2016

The acceleration is $10 \left(\frac{m}{s} ^ 2\right)$

Explanation:

Let the acceleration be $a$
Then we can use the equation $s = u t + \frac{1}{2} a {t}^{2}$
$u = 0$ initial velocity, starting frm rest
$s = 45 m$ distance
$t = 3 s$ time
So $45 = 0 + \frac{1}{2} a \cdot 9$
Aussuming that the downward direction is positive
$\frac{9}{2} a = 45$
So $a = 2 \cdot \frac{45}{9} = 2 \cdot 5 = 10 \left(\frac{m}{s} ^ 2\right)$
This is $g$