# Kinematic problem

by EvilBunny
Tags: kinematic
 P: 1,874 So for the first path you should get that $$t_1 = \frac{\sqrt{2gy_{high}}}{g}$$ For the second path you should use the distance equation $$t_2 = \sqrt{\frac{2 y_{net}}{g}}$$ These two times will give you just about everything. Is this what you got?