A test rocket is launched, starting on the ground, from rest, by accelerating it along an incline with constant acceleration a. The incline has length L, and it rises at theta degrees above the horizontal. At the instant the rocket leaves the incline, its engines turn off and it is subject only to gravity, g. Air resistance can be ignored. Taking the usual x-y coordinate system, with an origin at the top edge of the incline, a) what is the position vector when the rocket is at its highest point? b) what is the position vector when the rocket is on its way back down and once again at the same height as the top edge of the incline? Symbolic Answer.
The Attempt at a Solution
My question is this: Is it safe to assume that once you find the time it took for the rocket to reach it's highest point, that you can simply multiply it by 2 to find the time it took to reach y=o again?
I assume this is a yes, but what is throwing my thinking of is that the rocket did have an initial velocity in the y-direction, and when the rocket is at the top of the motion, the rocket no longer has that initial y-velocity coming down, so the time may be skewed.