Solve the Daring 510-N Swimmer's Minimum Speed Physics Problem"

AI Thread Summary
To solve the physics problem of a 510-N swimmer diving off a cliff, the key is to determine the minimum horizontal speed required to clear a 1.75 m wide ledge located 9.00 m below. The swimmer is treated as a projectile launched horizontally from a height of 9 meters. The vertical motion can be described using the equation y = y_0 + v_0t + 0.5at^2, while the horizontal distance is calculated using x = u cos(θ) t. The swimmer's mass does not affect the calculations, as only the forces acting on her during the fall are considered. The solution involves applying these equations and performing algebraic manipulations to find the required speed.
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A daring 510-N swimmer dives off a cliff with a running horizontal leap. What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 m wide and 9.00 m below the top of the cliff?
 
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You can find the time by using y=y_0+v_0t+.5at^2. That should get you started...
 
Consider her to be a projectile who is launching at 0 degrees to the horizontal and at a height of 9 metres. i don't think, ( and i jus looked at the problem offhandedly) that the mass of the diver comes into play.
now write equations for the vertical height and the horizontal. Msg me or post if u get stuck...after this, its just simple algebra.

RELATED EQUATIONS:

y = u \sin \theta - \frac{1}{2} g t^2

x = u \cos \theta t
 
smooth sid. msg me or post me...
 
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