A water balloon is launched at a speed of 42 m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located 27 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

u = 42 cos 36 horizontal speed
so
t = 27/(42 cos 36)

Vi = 42 sin 36 initial speed up

h = Vi t - 4.9 t^2