A water balloon is launched at a speed of 23 m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located 20 m from the launch pad. At what height above the ground level will the water balloon hit the building? Calculate the answer in meters (m) and rounded to three significant figures.

Answer 1

Horizontal problem:
u = 23 cos 36 until it stops

20 meters = u t
so
t = 20 /(23 cos 36)
use that t in the vertical problem

now vertical problem:

Vi = 23 sin 36 initial speed up

h = 0 + Vi t - 4.9 t^2