# Minimum speed physics problem

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?

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 dont 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....