A ball is shot from the top of a building with an initial velocity of 20 m/s at an angle θ = 39° above the horizontal. If a nearby building is the same height and 55 m away, how far below the top of the building will the ball strike the nearby building?

Answer 1

the horizontal speed is constant: 20 cos39°

The vertical position is
(20 sin39°)t - 4.9t^2

so, figure how long it takes to go 55m, then use that to get the relative height at that time.