A ball is launched off a ledge with an initial velocity of 14.0 m/s [E]. When it hits the ground, the angle of elevation from its landing point to the original launch position is 60 degrees. What are the vertical and horizontal displacements the ball?

Question

Thanks for any help! I've been stuck on this practice problem for a while now!

Answer 1

I assume that "14.0 m/s [E]" means the ball was launched horizontally.
So, if the initial height is h, then the distance traveled horizontally is h cos60° = h/2
the time taken to fall is t where
4.9t^2 = h
14t = h/2
so
4.9t^2 = 28t
t = 5.71 seconds
Now you can find h and h/2

Answer 2

Thanks so much oobleck! That really helped.