Question

A grapefruit falls from a tree and hits the ground .75 seconds later. How far did the grapefruit drop? What was its speed?

Answer 1

1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration
`g=9.81 m/s^2`. Calling h its height at t=0, the height at time t is given by

`h(t)=h- (1)/(2)gt^2`
We are told thatn when
`t=0.75 s` the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:

`0=h- (1)/(2)gt^2`

`h= (1)/(2)gt^2= (1)/(2)(9.81 m/s^2)(0.75 s)^2=2.76 m`

2) The speed of the grapefruit at time t is given by

`v(t)=v_0 +gt`
where
`v_0=0` is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:

`v(0.75 s)=gt=(9.81 m/s^2)(0.75 s)=7.36 m/s`