A cat jumps off of the table with a velocity vi at an angle θ. The table is h meters tall. How far away from the table is the cat?
V(x) = ViCosθ
V(y) = ViSinθ
I first solved for time using 0 = Xi + Vit +1/2a(t^2)
The Attempt at a Solution
I got time as
t = ViSinθ-√(Vi2Sinθ2 + 2gh) / g
But my teacher got a answer slightly different. Am I solving for time the write why? I think so. If not, What am I missing? Thanks