A solid, uniform ball rolls without slipping up a hill. At the top of the hill it is moving horizontally; then it goes over the vertical cliff. V= 29 m/s and H= 22.0m A) How far from the foot of the cliff does the ball land? B) How fast is it moving just before it lands? Been stuck on this one for a while but this is what I have so far. From my understanding before the ball starts rolling up the hill it has pure kinetic energy. As the ball goes up the hill, kinetic energy gets transferred to potential, but at the top of the hill it will still have some kinetic energy and the rest stored as potential energy. When the ball flies off the cliff it will act as a projectile and have purely potential energy. I broke the problem down into thinking that I need to find the velocity of the ball before it flies off the cliff so I did an energy balance on the ball before it goes up the hill and after it goes up the hill: mgh = (.5mvf2 -.5mvi2) + (.5Iωf2-.5Iωi2) Then solved for vf: √(gh+7/10*vi2)/(7/10) = 33.8969 m/s I then used the projectile motion equation y = y0 + v0 - .5gt2 Using the quadratic formula I obtained the time that it took 7.51 seconds to fall back to the ground. I substituted this time value into the projectile motion equation for x and obtained 255 m which seems way too high. I'm kind of at a loss as to what to do with this problem. I feel like my approach is correct but the error is coming from the energy balance, I don't think I'm including something that needs to be there. A nudge in the right direction would be very much appreciated.