Question

A seaplane flies horizontally over the ocean at 50 meters/second. It releases a buoy, which lands after 21 seconds. What's the vertical displacement from the plane to the ocean? (Note: Displacement will end up negative, since the direction is down.)

Answer 1

The motion of the buoy is a composition of two independent motions:
- a uniform motion on the horizontal axis, with constant speed vx=50 m/s
- an uniformly accelerated motion on the vertical axis, with constant acceleration $g=9.81 m/s^2$

Since we want to find the vertical displacement, we are only interested in the vertical motion.
The law of motion on the vertical direction is given by:
$y(t)=h+v_{0y} t+ \frac{1}{2}gt^2$
where
h is the initial height of the buoy
$v_{0y}$ is the initial vertical velocity of the buoy, which is zero
t is the time

We know that the buoy lands after t=21 seconds, this means that the vertical position at t=21 s is y(21 s)=0. If we substitute these data into the equation, we can find the value of h, the initial height of the buoy:
$0=h+ \frac{1}{2}gt^2$
$h= -\frac{1}{2}gt^2= -\frac{1}{2}(9.81 m/s^2)(21 s)^2=-2163 m$
And this corresponds to the vertical displacement of the buoy.