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

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SUMMARY

The minimum pushoff speed required for cliff divers in Acapulco to clear rocky outcrops is determined by analyzing the motion equations for projectile motion. Given a height of 35 m and a horizontal distance of 5 m, the time in the air can be calculated using the equation Yf = Yi + Vt + 0.5(9.81)t^2. By setting the initial vertical velocity to zero and solving for time, the horizontal velocity can then be derived as the distance divided by the time calculated.

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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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