Uneven start/end points (2D Kinematics)

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SUMMARY

The discussion focuses on solving a 2D kinematics problem involving a diver who springs from a board 3 meters above water, reaching a speed of 9.30 m/s at an angle of 84.0º upon contact with the water. The key equations utilized include h = v(i) * t - 1/2at^2, t(half) = (v * sin(x))/g, and d = (v^2 * 2sin(x) * cos(x))/g. The solution was achieved by applying the quadratic formula to isolate the time variable, ultimately determining the diver's initial velocity and trajectory.

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Homework Statement


A diver springs upward from a board that is 3 meters above the water. At the instant she contacts the water her speed is 9.30 m/s and her body makes an angle of 84.0º with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.

Homework Equations


h = v(i) * t - 1/2at^2
t(half) = (v * sin(x))/g
d = (v^2 * 2sin(x) * cos(x))/g

The Attempt at a Solution


Tried using h = v(i) * t - 1/2at^2 to find the initial velocity (with h as 3).
Unsure exactly of where I should be headed...

*SOLVED*
 
Last edited:
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Using h = vt - 1/2at^2, I've gotten the quadratic (-4.9t^2 + 9.249t + 3). From there, to determine the velocity (magnitude), I isolated the portion of the parabola between y = 3 (x = 1.888) and y = 0 (x = 2.17). I'm just unsure of how to find the time with the diver starting 3 meters above the water.
 

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