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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.

2 Answers

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Answer:

The answer for part A is: y

For part B it is :
y=y0 + v y0 ^t +(1)/(2) ay^t^2

Explanation: edmentum sample answer

User Ayman Elshehawy
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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- (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- (1)/(2)gt^2

h= (1)/(2)gt^2= (1)/(2)(9.81 m/s^2)(21 s)^2=2163 m
User Bob Uni
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