A ball with mass m rolls over the edge of a table h meters above the ground. It lands a distance x meters from a point directly below the edge of the table. Show that the speed of the ball at the edge of the table was v=x/sq rt 2h/g and that the kinetic energy of the ball immediately before it hits the ground is k=mg(x^2 + 4h^2)/4h

1 answer

sqrt(2h/g) is the time it takes to fall. Dividing x by that gives you the horizontal speed Vx at which the ball leaves the table.

Use Vx and Vy to compute the total KE.
Vy = g t.