Some enterprising physics students working
on a catapult decide to have a water balloon
fight in the school hallway. The ceiling is of
height 3.2 m, and the balloons are launched
at a velocity of 12 m/s.
The acceleration of gravity is 9.8 m/s2 .
At what angle must they be launched to
just graze the ceiling?
Answer in units of �.
To determine the launch angle at which the water balloon just grazes the ceiling, we can use the equations of projectile motion.
Let's break down the problem:
Given:
- Height of the ceiling, h = 3.2 m
- Initial velocity of the water balloon, v = 12 m/s
- Acceleration due to gravity, g = 9.8 m/s^2
We need to find the launch angle, θ.
We can use the equation for the maximum height reached by a projectile:
h = (v^2 * sin^2(θ)) / (2 * g)
Rearranging the equation:
sin^2(θ) = (2 * g * h) / v^2
Taking the square root of both sides:
sin(θ) = sqrt((2 * g * h) / v^2)
Now, we can solve for θ by taking the inverse sine (arcsine) of both sides:
θ = arcsin(sqrt((2 * g * h) / v^2))
Plugging in the given values:
θ = arcsin(sqrt((2 * 9.8 * 3.2) / 12^2))
Calculating the expression inside the sqrt:
θ = arcsin(sqrt(0.51733))
Taking the square root:
θ = arcsin(0.719)
Using a calculator or trigonometric table, we find that the arcsin(0.719) is approximately 45.47 degrees.
Therefore, the water balloon must be launched at an angle of approximately 45.47 degrees to just graze the ceiling.