A 1.60 kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 195 N/m and a 280 g metal ball is in the tray. The spring is below the tray, so it can oscillate up-and-down. The tray is then pushed down 15.0 cm below its equilibrium point (call this point A) and released from rest.
a) How high above point A will the tray be when the metal ball leaves the tray? (Hint: This does not occur when the ball and tray reach their maximum speeds.)
b) How much time elapses between releasing the system at point A and the ball leaving the tray?
c) How fast is the ball moving just as it leaves the tray?
The Attempt at a Solution
So I already solved part (a) by equating the accelerations of the ball and the spring system, g=xω2 , and got an answer for x. But I can't seem to figure out how to get the time at this instant for part (b): I tried using x=Acos(ωt+[itex]\phi[/itex]) and solving for t, but it doesn't make too much sense to me. And I believe I can solve for part (c) with the time t from part (b) using v=-ωAsin(ωt+[itex]\phi[/itex]), but this I'm also not 100 percent on.
Could anyone nudge me in the right direction? I seem to be stuck after part (a)