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°

Answer: 50° above the horizontal