The motion of the buoy consists of two independent motions on the horizontal and vertical axis.
On the horizontal axis, the motion of the buoy is a uniform motion with constant speed [tex]v=50 m/s[/tex]. On the vertical axis, the motion of the buoy is a uniformly accelerated motion with constant acceleration [tex]g=9.81 m/s^2[/tex]. The vertical position of the buoy at time t is given by
[tex]y(t)=h- \frac{1}{2}gt^2 [/tex]
where h is the initial heigth of the buoy when it is released from the plane. At the time t=21 s, the buoy reaches the ground, so y(21 s)=0. If we substitute these two numbers inside the equation, we can find the value of h, the vertical displacement from the plane to the ocean:
[tex]0=h- \frac{1}{2}gt^2 [/tex]
[tex]h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(21 s)^2=2163 m [/tex]
