Question

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 34.0 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 1.90 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.

Answer 1

(a) The first balloon travels for 1.90 s. Since we know both the horizontal distance and the balloon's flight time, we know its horizontal velocity:

vx = dx / dt = 34.5m / 1.90 s = 18.16 m/s horizontally

Next we must find the initial vertical velocity. We know the downward acceleration is -9.81 m/s², We also know it was launched from an initial height of 15.0 m and impacted 11.0 m above the ground (half of Walton dorm's height). That means its overall vertical displacement was -4.0 m. Given the flight time of 1.90 s, we can work out the initial vertical velocity using the formula:

dy = vy*t + 1/2 at²

-4.0 m = vy(1.90 s) + 1/2(-9.81 m/s²)(1.90 s)²

-4.0 m = vy(1.90 s) - 17.70 m

vy(1.90s) = 13.70m

vy = 7.21 m/s vertically

So the initial horizontal velocity was 18.16 m/s and the initial vertical velocity was 7.21 m/s. These velocities represent the opposite (7.21 m/s) and adjacent (18.16m/s) sides of a right triangle whose hypotenuse is the launch speed, so we can use the tangent function to calculate the angle:

tanθ = opp / adj = (7.21 m/s) / (18.16m/s) = 0.40

tanθ = 0.40

θ = arctan(0.40) = 21.67º above the horizontal. ....Answer.

(b) We know the angle of launch, 21.67º, but not what speed it must have in order to reach its target. This is how I would visualize the problem: without gravity the water balloon will fly off in a straight line at an angle of 21.67º above the horizontal. It would miss the top of Waldon by some distance y which we will determine using a little bit of trigonometry. We will then determine how long it would take gravity to pull the water balloon down by that amount. That is the balloon's flight time. Given the flight time we can work out the horizontal velocity. Since we know the ratio between the horizontal and vertical velocity, we can then work out the vertical velocity, and knowing both of those, we can work out the launch velocity.

If the balloon is launched at an angle of 21.67º and travels in a straight line, by the time it has moved 34m horizontally it will be 13.6 meters above its launch point:

dy = dx*tan(21.67º)

dy = 34m * 0.40

dy = 13.6 m

The difference between the height of the two dorms is only 7.0 meters, so if unaffected by gravity, the water balloon would pass the edge of Waldon dorm 6.6 meters above it. Gravity has to pull the balloon downward by 6.6 meters during its flight time. How long does gravity take to pull an object down by 6.6 meters?

This question can be answered using the formula:

dy = 1/2 at²

-6.6 m = 1/2 (-9.81 m/s²)t²

-6.6 m = (-4.905 m/s²)t²

t² = 1.345 s²

t = 1.6 s

Whatever the balloon's initial speed is, its flight time is 1.6 seconds. We can now calculate the horizontal component of the balloon's initial velocity:

vx = dx / dt

vx = (34m) / (1.6 s) = 21.25 m/s

And now the vertical component. Since we know the flight time, we can find the vertical component the same way we did before:

dy = vy*t + 1/2at²

The dy in this case is the vertical displacement of the balloon from the start of its flight to the end. We no longer need to worry about the hypothetical displacement we used earlier.

7.0 m = vy*(1.6s) + 1/2(-9.81 m/s²)(1.6 s)²

7.0 m = vy(1.6 s) - 12.55 m

19.55m = vy(1.6 s)

vy = 12.22 m/s

Now that we know the horizontal and vertical components of the initial launch velocity, we can use the Pythagorean theorem to find the velocity itself

vo² = vx² + vy²

vo² = (21.25m/s)² + (12.22 m/s)² = 600.96 m²/s²

vo = 24.5 m/s....Answer.

Hope this helps you.