In the annual battle of the dorms, students gather on the roofs of Jackson and Walton dorms to launch water balloons at each other with slingshots. The horizontal distance between the buildings is 36.0 m, and the heights of the Jackson and Walton buildings are, respectively, 15.0 m and 22.0 m. The first balloon launched by the Jackson team hits Walton dorm 2.20 s after launch, striking it halfway between the ground and the roof. Ignore air resistance. (a) Find the direction of the balloon's initial velocity. Give your answer as an angle (in degrees) measured above the horizontal. (b) A second balloon launched at the same angle hits the edge of Walton's roof. Find the initial speed of this second balloon
To find the direction of the balloon's initial velocity, you can use the kinematic equations of motion.
First, let's analyze the vertical motion of the balloon. We know that the balloon hits halfway between the ground and the roof of Walton dorm, which means it travels a vertical distance of (22.0 m + 0 m)/2 = 11.0 m. We can use the equation for vertical displacement in free-fall motion:
y = v0y * t + (1/2) * g * t^2
where y is the vertical displacement, v0y is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the known values, we have:
11.0 m = v0y * 2.20 s + (1/2) * 9.8 m/s^2 * (2.20 s)^2
Simplifying the equation, we can solve for v0y:
11.0 m = v0y * 2.20 s + 10.034 m
v0y * 2.20 s = 11.0 m - 10.034 m
v0y * 2.20 s = 0.966 m
v0y = 0.966 m / 2.20 s
v0y ≈ 0.439 m/s
Now, let's analyze the horizontal motion of the balloon. The horizontal distance between the buildings is 36.0 m, and the time of flight is 2.20 s. We can use the equation for horizontal displacement:
x = v0x * t
where x is the horizontal displacement and v0x is the initial horizontal velocity.
Substituting the known values, we have:
36.0 m = v0x * 2.20 s
v0x = 36.0 m / 2.20 s
v0x ≈ 16.4 m/s
To find the direction of the balloon's initial velocity, we can use trigonometry.
The angle θ (measured above the horizontal) can be found using the tangent function:
tan(θ) = v0y / v0x
θ = arctan(v0y / v0x)
θ = arctan(0.439 m/s / 16.4 m/s)
θ ≈ 1.54 degrees (rounded to two decimal places)
Therefore, the direction of the balloon's initial velocity is approximately 1.54 degrees above the horizontal.
Now, to find the initial speed of the second balloon that hits the edge of Walton's roof at the same angle, we can use the range formula for projectile motion:
R = (v0^2 * sin(2θ)) / g
where R is the range, v0 is the initial speed, θ is the launch angle (1.54 degrees), and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the known values, we have:
36.0 m = (v0^2 * sin(2 * 1.54 degrees)) / 9.8 m/s^2
Rearranging the equation, we can solve for v0:
v0^2 = (36.0 m * 9.8 m/s^2) / sin(2 * 1.54 degrees)
v0 ≈ √((36.0 m * 9.8 m/s^2) / sin(2 * 1.54 degrees))
v0 ≈ 16.9 m/s (rounded to two decimal places)
Therefore, the initial speed of the second balloon is approximately 16.9 m/s.