Question

A grapefruit falls from a tree and hits the ground 0.89 s later. how far did the grapefruit drop? what was its speed when it hit the ground?

Answer 1

1) The free fall motion of the grapefruit is an uniformly accelerated motion, with constant acceleration equal to $g=9.81 m/s^2$, therefore the distance covered by the grapefruit in a time t is equal to
$S= \frac{1}{2}gt^2$
Substituting $t=0.89 s$, we get how far the grapefruit dropped:
$S= \frac{1}{2}(9.81 m/s^2)(0.89s)^2=3.89 m$

2) The speed at time t in an uniformly accelerated motion of free fall is given by
$v(t)=v_0 + gt$
where $v_0$ is the initial speed, that in this case was zero. Therefore, using t=0.89 s we can find the speed just before the grapefruit hits the ground:
$v(0.89 s)=gt=(9.81 m/s^2)(0.89s)=8.7 m/s$