A marble is rolling across a smooth 1.2 m tall table at a velocity of 3 m/s. How far from the edge of the table does it land? (SHOW ALL OF YOUR WORK)

S = 1/2 g t^2      where t is the time to fall 1.2 m

t = (2 S / g)^1/2 = (2 * 1.2 / 9.8) = .495 s

Sy = Vy T = 3 m/sec  * .495  sec = 1.48 m     distance from edge of table

(Rotational speed has no effect since table is smooth)

Gravitational potential energy = (mass) x (gravity) x (height)

GPE = (4 kg) x (9.8 m/s²) x (3 m)

GPE = (4 x 9.8 x 3) kg-m²/s²

GPE = 117.6 Joules