Motion Problem (2-D): Find Pushoff Speed & Time in Air

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To determine the minimum pushoff speed for cliff divers in Acapulco, the vertical motion equation is used to calculate the time spent in the air. The divers must clear a horizontal distance of 5.0 m while falling from a height of 35 m. By setting the initial vertical velocity and position to zero, the time of flight can be derived from the vertical motion equation. Once the time is established, the horizontal pushoff speed can be calculated by dividing the horizontal distance by the time. This approach effectively relates the x and y components of motion to solve the problem.
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Homework Statement



The cliff divers of Acapulco push off horizontally from rock platforms about 35 m above the water, but they must clear rocky outcrops at water level that extend out into the water 5.0 m from the base of the cliff directly under their launch point.

What minimum pushoff speed is necessary to clear the rocks?
How long are they in the air?

Homework Equations



Yf = Yi + Vt + 0.5(9.81)t^2

The Attempt at a Solution



I'm having trouble relating time to the x & y components because I cannot see how the velocity is found.
 
Last edited:
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I found it out!

The time can be found by using the y-component equation, setting the velocity & initial y-position both to 0. We know the x-distance traveled is 5 m, so the velocity is 5 m divided by the value for time that was found.
 

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