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= (1)/(2)gt^2`
Substituting
`t=0.89 s`, we get how far the grapefruit dropped:

`S= (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`