1) The free fall motion of the grapefruit is an uniformly accelerated motion, with constant acceleration equal to [tex]g=9.81 m/s^2[/tex], therefore the distance covered by the grapefruit in a time t is equal to
[tex]S= \frac{1}{2}gt^2 [/tex]
Substituting [tex]t=0.89 s[/tex], we get how far the grapefruit dropped:
[tex]S= \frac{1}{2}(9.81 m/s^2)(0.89s)^2=3.89 m [/tex]
2) The speed at time t in an uniformly accelerated motion of free fall is given by
[tex]v(t)=v_0 + gt[/tex]
where [tex]v_0[/tex] 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:
[tex]v(0.89 s)=gt=(9.81 m/s^2)(0.89s)=8.7 m/s[/tex]
