Some enterprising physics students working on a catapult decide to have a water balloon fight in the school hallway. the ceiling is of height 3.6 m, and the balloons are launched at a velocity of 11 m/s. the acceleration of gravity is 9.8 m/s 2 . at what angle must they be launched to just graze the ceiling?

Given:
v = 11 m/s, the launch speed
h = 3.6 m, the height of the ceiling

Assume g = 9.8 m/s², and neglect air resistance.
Let θ =  the launch angle, measured above the horizontal.
The initial vertical speed is
u = 11 sin θ m/s 

At the maximum height of 3.6 m, the vertical velocity is zero.
Therefore
(11 sinθ m/s)² - 2*(9.8 m/s²)*(3.6 m) = 0
121 sin² θ = 70.56
sin²θ = 0.5831
sin θ = 0.7636
θ = sin⁻¹ 0.7636 = 49.8° ≈ 50°

50° above the horizontal