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