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
$h(t)=h- \frac{1}{2}gt^2$
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:
$0=h- \frac{1}{2}gt^2$
$h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(0.75 s)^2=2.76 m$

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