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.) Part A: Enter the variable symbol for the quantity you need to find. Use your keyboard and the keypad to enter your answer. Then click Done.

Answer 1

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 $v=50 m/s$. On the vertical axis, the motion of the buoy is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$. The vertical position of the buoy at time t is given by
$y(t)=h- \frac{1}{2}gt^2$
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:
$0=h- \frac{1}{2}gt^2$
$h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(21 s)^2=2163 m$