1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration [tex]g=9.81 m/s^2[/tex]. Calling h its height at t=0, the height at time t is given by
[tex]h(t)=h- \frac{1}{2}gt^2 [/tex]
We are told thatn when [tex]t=0.75 s[/tex] 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:
[tex]0=h- \frac{1}{2}gt^2 [/tex]
[tex]h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(0.75 s)^2=2.76 m [/tex]
2) The speed of the grapefruit at time t is given by
[tex]v(t)=v_0 +gt[/tex]
where [tex]v_0=0[/tex] is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
[tex]v(0.75 s)=gt=(9.81 m/s^2)(0.75 s)=7.36 m/s[/tex]
