A solid sphere of diameter 0.15m and mass 0.5kg is released and rolls without slipping down the ramp. The ramp is 0.81m high and is on a table. The table has a height of 1.40 meters. The balls move freely once it leaves the table. a) What is the moment of inertia of the ball about the axis of rotation? I = (2/5)(0.5kg)(0.075)^2 = 0.001125 kg*m^2 b) Find the angular speed of the ball when it reaches the table top. Potential energy = Kinetic Energy (0.5)(9.81)(0.81) = (1/2)(0.001125)w^2 ---> w=84.04 rad/sec. c) Find the initial velocity (magnitude and direction) of the ball when it leaves the table. Total Energy = 0 = (1/2)(0.5)(v)^2 - (1/2)(0.001125)(84.04)^2 velocity = 3.987m/s How do I find the direction? d) Find the distance d (distance ball travels when it left the table). x = vt + (1/2)at^2 -1.40y + dx = (3.987x)t + (-4.905y)t^2 x: d = 3.987t y: -1.40 = -4.905t^2 ----> t=0.534sec. d=3.987(0.534) ----> d=2.13meters Are my answers correct? Did I do anything wrong? Please help. Thanks!