1. Velocity at which the packet reaches the ground: 121.2 m/s
The motion of the packet is a uniformly accelerated motion, with constant acceleration [tex]a=g=9.8 m/s^2[/tex] directed downward, initial vertical position [tex]d=750 m[/tex], and initial vertical velocity [tex]v_0 = 0[/tex]. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:
[tex]v_f^2 -v_i^2 =2ad[/tex]
substituting, we find
[tex]v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s[/tex]
2. height at which the packet has half this velocity: 562.6 m
We need to find the heigth at which the packet has a velocity of
[tex]v_f=\frac{121.2 m/s}{2}=60.6 m/s[/tex]
In order to do that, we use again the same SUVAT equation substituting [tex]v_f[/tex] with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:
[tex]v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m[/tex]
Which means that the heigth of the packet was
[tex]h=750 m-187.4 m=562.6 m[/tex]
