A water balloon is launched at a building 24m away with an initial velocity of 18 m/s at an angle of 50° above the horizontal. At what height will the balloon strike the building?

Answer 1

First, we can find the time it takes for the projectile (the water balloon) to reach the building by using the horizontal component of the motion.

The horizontal component of the velocity can be found using the formula:

v_x = v * cos(θ)

v_x = 18 m/s * cos(50°) ≈ 11.58 m/s

Now, we can find the time it takes to reach the building by dividing the distance by the horizontal component of the velocity:

t = (distance) / (horizontal velocity)
t = 24 m / 11.58 m/s ≈ 2.07 s

Now that we know the time it takes to reach the building, we can find the height at which the balloon strikes the building using the vertical component of the motion.

The vertical component of the velocity can be found using the formula:

v_y_initial = v * sin(θ)

v_y_initial = 18 m/s * sin(50°) ≈ 13.79 m/s

Now we can calculate the height using the equation of motion for the vertical displacement (assuming that the initial height is 0):

y = v_y_initial * t - 0.5 * g * t^2

where g = 9.81 m/s^2 (acceleration due to gravity)

y = 13.79 m/s * 2.07s - 0.5 * 9.81 m/s^2 * (2.07s)^2
y ≈ 28.53 m - 0.5 * 9.81 m/s^2 * 4.28 s^2
y ≈ 28.53 m - 21.01 m
y ≈ 7.52 m

So, the height at which the balloon strikes the building is approximately 7.52 meters.