Could anyone please verify with me that I have the right idea in answering the question below. 1. The problem statement, all variables and given/known data If you know the initial velocity v0 and the initial and final heights y0 and y, you can use x=x0+v0t+(1/2)at^2 to solve for the time t when the object will be at height y. But the equation is quadratic in t, so you'll get two answers. Physically, why is this? 3. The attempt at a solution This occurs because during the motion of an object it has passed through it's initial displacement on 2 occassions. Thus times occur twice.