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 40.0 m, and the heights of the Jackson and Walton buildings are, respectively, 14.0 m and 22.0 m. The first balloon launched by the Jackson team hits Walton dorm 2.00 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.

(a) Ah, the annual battle of the dorms, where water balloons fly high and friendships get soaked! To find the direction of the balloon's initial velocity, we can use some projectile motion calculations.

Let's consider the horizontal and vertical components of the balloon's motion separately.

In the vertical direction, we know that the balloon hits Walton dorm halfway between the ground and the roof. Since the height of the Walton building is 22.0 m, the balloon reaches a maximum height of 11.0 m.

The time taken to reach the maximum height can be found using the equation:
t = sqrt(2 * h / g)
where t is the time, h is the height, and g is the acceleration due to gravity. Plugging in our values, we get:
t = sqrt(2 * 11.0 / 9.8) ≈ 1.48 s

Since the total time in the air is 2.00 s, the time taken to reach the maximum height and come back down must be 2.00 s - 1.48 s = 0.52 s.

Now, let's look at the horizontal direction. We know the distance between the buildings is 40.0 m. The horizontal velocity can be found using the equation:
v = d / t
where v is the horizontal velocity, d is the distance, and t is the time. Plugging in our values, we get:
v = 40.0 / 2.00 ≈ 20.0 m/s

Since the initial velocity has both horizontal and vertical components, we can use trigonometry to find the angle.

tan(theta) = v(vertical) / v(horizontal)
tan(theta) = (11.0 / 0.52) / 20.0
tan(theta) ≈ 1.058

Taking the arctan of both sides, we find that theta ≈ 47.5 degrees.

So, the direction of the balloon's initial velocity is approximately 47.5 degrees above the horizontal.

(b) Ah, the second balloon wants to make it to the roof's edge! To find its initial speed, we need to consider the horizontal and vertical components of the motion again.

Since the horizontal velocity remains the same as in part (a), which was 20.0 m/s, we can use the given distance and time of flight to find the vertical velocity.

The vertical distance is 22.0 m, and the total time in the air is 2.00 seconds.

Using the equation:
d = v(initial) * t + (1/2) * a * t^2
where d is the vertical distance, v(initial) is the initial vertical velocity, t is the time, and a is the acceleration due to gravity (-9.8 m/s^2).

Plugging in our values, we get:
22.0 = v(initial) * 2.00 + (1/2) * (-9.8) * 2.00^2
22.0 = 2.00 * v(initial) - 19.6
2.00 * v(initial) = 22.0 + 19.6
2.00 * v(initial) = 41.6
v(initial) ≈ 20.8 m/s

Therefore, the initial speed of the second balloon is approximately 20.8 m/s.

So, there you have it! The direction of the first balloon's initial velocity is approximately 47.5 degrees above the horizontal, and the initial speed of the second balloon is approximately 20.8 m/s. Good luck to all the water balloon warriors in their battle of the dorms!

To solve this problem, we can use the equations of motion for projectiles. Let's solve the problem step by step:

Step 1: Find the direction of the balloon's initial velocity.
Given:
- Horizontal distance between the buildings (range): 40.0 m
- Height of Jackson dorm: 14.0 m
- Height of Walton dorm: 22.0 m
- Time of flight (total time taken by the balloon to reach the target): 2.00 s

Let's assume the angle above the horizontal at which the balloon was launched is θ.

Using the equation for vertical displacement, we have:
Δy = v₀sin(θ)t + (1/2)gt²
where Δy is the vertical displacement, v₀ is the initial velocity, θ is the angle above the horizontal, t is the time of flight (2.00 s), and g is the acceleration due to gravity (-9.8 m/s²).

For the balloon to hit halfway between the ground and the roof of Walton dorm, the vertical displacement is (22 m - 14 m)/2 = 4.0 m.

Substituting the known values into the equation and solving for θ, we get:
4.0 = v₀sin(θ)(2.00) + (1/2)(-9.8)(2.00)²
4.0 = 2.00v₀sin(θ) - 19.6

Rearranging the equation, we have:
2.00v₀sin(θ) = 23.6

Dividing both sides by 2.00, we get:
v₀sin(θ) = 11.8

Since sin(θ) = opposite/hypotenuse, and the opposite side is the vertical component of the initial velocity (v₀sin(θ)), and the hypotenuse is the initial velocity (v₀), we can rewrite the equation as:
sin(θ) = 11.8/v₀

Taking the inverse sine of both sides, we get:
θ = arcsin(11.8/v₀)

Therefore, the direction of the balloon's initial velocity is θ = arcsin(11.8/v₀).

Step 2: Find the initial speed of the second balloon.
Given:
- Horizontal distance between the buildings (range): 40.0 m
- Angle above the horizontal (θ): the same angle as in Step 1

To find the initial speed (v₀) of the second balloon, we need to use the equation for range:
R = v₀cos(θ)t
where R is the horizontal distance, v₀ is the initial velocity, θ is the angle above the horizontal (same as in Step 1), and t is the time of flight (2.00 s).

Substituting the known values and solving for v₀, we have:
40.0 = v₀cos(θ)(2.00)

Rearranging the equation, we get:
v₀cos(θ) = 20.0

Dividing both sides by cos(θ), we get:
v₀ = 20.0/cos(θ)

We already found θ in Step 1 as θ = arcsin(11.8/v₀). Substitute this value into the equation, and we have:
v₀ = 20.0/cos(arcsin(11.8/v₀))

To solve this equation, we can use numerical methods or approximation techniques. Let's assume v₀ ≈ 24.0 m/s.

Now, we can substitute this value of v₀ into the equation:
v₀ = 20.0/cos(arcsin(11.8/24.0))

Using a calculator, we find v₀ ≈ 23.4 m/s (rounded to three significant figures).

Therefore, the initial speed of the second balloon is approximately 23.4 m/s.

To solve this problem, we can use the equations of motion in projectile motion. Let's break the problem down into parts.

(a) To find the direction of the balloon's initial velocity, we need to determine the launch angle. We know the horizontal distance between the buildings (40.0 m), the height of the Jackson building (14.0 m), and the time it takes for the balloon to reach the Walton dorm (2.00 s).

1. Determine the horizontal velocity: Since there is no horizontal acceleration, the horizontal velocity remains constant. We can use the formula:

Horizontal distance = Horizontal velocity × Time
40.0 m = Horizontal velocity × 2.00 s

Solve for the horizontal velocity:
Horizontal velocity = 40.0 m / 2.00 s = 20.0 m/s

2. Determine the vertical velocity: We know the initial vertical position (halfway between the ground and the Walton dorm's roof) and the time it takes for the balloon to reach the Walton dorm (2.00 s). We can use the formula:

Vertical distance = Initial vertical velocity × Time + (1/2) × Acceleration × Time²
22.0 m = Initial vertical velocity × 2.00 s + (1/2) × (-9.8 m/s²) × (2.00 s)²

Solve for the initial vertical velocity:
Initial vertical velocity = (22.0 m - 1/2 × (-9.8 m/s²) × (2.00 s)²) / 2.00 s = 9.4 m/s

3. Find the magnitude of the initial velocity: The initial velocity is the vector sum of the horizontal and vertical velocities. We can use the Pythagorean theorem:

Initial velocity = √(Horizontal velocity² + Initial vertical velocity²)
Initial velocity = √(20.0 m/s)² + (9.4 m/s)²)
Initial velocity = √(400 m²/s² + 88.36 m²/s²)
Initial velocity = √(488.36 m²/s²)
Initial velocity ≈ 22.10 m/s

4. Find the launch angle: The launch angle is the angle measured above the horizontal. We can use the inverse tangent function:

Launch angle = arctan(Initial vertical velocity / Horizontal velocity)
Launch angle = arctan(9.4 m/s / 20.0 m/s)
Launch angle ≈ arctan(0.47)
Launch angle ≈ 25.77°

Therefore, the balloon's initial velocity is at an angle of approximately 25.77° measured above the horizontal.

(b) To find the initial speed of the second balloon that hits the edge of Walton's roof, we can use the same launch angle found in part (a) and the known horizontal distance (40.0 m).

1. Determine the vertical distance: Since the second balloon hits the edge of the roof, which is 22.0 m high, the vertical distance is 22.0 m.

2. Use the launch angle and the horizontal distance to find the initial speed: We can use the following formula for the range in projectile motion:

Range = (Initial speed)² × sin(2 × Launch angle) / gravitational acceleration
40.0 m = (Initial speed)² × sin(2 × 25.77°) / 9.8 m/s²

Solve for the initial speed:
(Initial speed)² = (40.0 m) × 9.8 m/s² / sin(2 × 25.77°)
Initial speed = √((40.0 m) × 9.8 m/s² / sin(2 × 25.77°))

Calculate the initial speed:
Initial speed ≈ √(391.2 m²/s² / sin(51.54°))
Initial speed ≈ √(391.2 m²/s² / 0.792)
Initial speed ≈ √495 m²/s²
Initial speed ≈ 22.25 m/s

Therefore, the initial speed of the second balloon launched at the same angle as the first balloon is approximately 22.25 m/s.

hi = 14

h = 22/2 = 11

u = 40m/2s = 20 m/s

11 = 14 + Vi (2) - 4.9 (4)
2 Vi = -3 + 19.6
Vi = 8.3 m/s

tan theta = Vi/u = 8.3/20
so theta = 22.5 degrees

Now you should be able to part b