A disgruntled physics student,frustrated with finals,releases his tensions by bombarding the adjacent building,17.7 m away with water balloons. He fires one at 38 degrees from the horizontal with an initial speed of 27.1 m/s. The acceleration of gravity is 9.8 m/s^2. for how long is the balloon in the air? Answer in seconds

How far above the initial launch height does the balloon hit the opposing building? Answer in m

To find the time for which the water balloon is in the air, we can use the equation for projectile motion:

y = y0 + v0y * t - (1/2)gt^2

where:
- y is the vertical displacement
- y0 is the initial launch height (assumed to be 0)
- v0y is the initial vertical velocity
- t is the time
- g is the acceleration due to gravity

Since the balloon lands on the opposing building, the final vertical displacement (y) will also be 0. So, the equation becomes:

0 = 0 + v0y * t - (1/2)gt^2

We can use the initial launch angle (θ = 38 degrees) to find the vertical component of initial velocity (v0y):

v0y = v0 * sin(θ)

Substituting the given values:

v0y = 27.1 m/s * sin(38 degrees) = 16.52 m/s

Now, we can rewrite the equation:

0 = 0 + 16.52 m/s * t - (1/2) * 9.8 m/s^2 * t^2

Rearranging the equation:

(1/2) * 9.8 m/s^2 * t^2 - 16.52 m/s * t = 0

Simplifying further:

4.9 m/s^2 * t^2 - 16.52 m/s * t = 0

We can factor out t:

t * (4.9 m/s^2 * t - 16.52 m/s) = 0

This equation yields two solutions: t = 0 and t = 3.38 seconds. Since time cannot be negative in this scenario, the time for which the balloon is in the air is:

t = 3.38 seconds

To find the vertical distance above the initial launch height where the balloon hits the opposing building, we can use the equation:

y = y0 + v0y * t - (1/2)gt^2

Substituting values:

y = 0 + 16.52 m/s * 3.38 seconds - (1/2) * 9.8 m/s^2 * (3.38 seconds)^2

Calculating:

y = 16.52 m/s * 3.38 seconds - (1/2) * 9.8 m/s^2 * (3.38 seconds)^2

y = 55.88 m - 54.22 m

y = 1.66 m

Therefore, the water balloon hits the opposing building approximately 1.66 meters above the initial launch height.

To find the time of flight and the vertical displacement of the water balloon, we can break the motion into horizontal and vertical components and use kinematic equations.

First, let's find the time of flight:

1. Split the initial velocity into horizontal and vertical components:
- Horizontal component: Vx = initial speed * cos(theta), where theta is the launch angle.
Vx = 27.1 m/s * cos(38 degrees).
- Vertical component: Vy = initial speed * sin(theta), where theta is the launch angle.
Vy = 27.1 m/s * sin(38 degrees).

2. Calculate the time of flight using the vertical component of velocity:
- The time for the object to reach its peak height and then fall back to the ground is twice the time it takes to reach the peak.
- t = 2 * (Vy / g), where g is the acceleration due to gravity.
t = 2 * (27.1 m/s * sin(38 degrees) / 9.8 m/s^2).

Next, let's find the vertical displacement of the balloon:

3. Determine the vertical displacement using the vertical component of velocity:
- The vertical displacement, h, can be calculated using the equation: h = (Vy^2) / (2 * g), where g is the acceleration due to gravity.
h = (27.1 m/s * sin(38 degrees))^2 / (2 * 9.8 m/s^2).

Now let's calculate the values:

1. Calculate the horizontal component of velocity:
Vx = 27.1 m/s * cos(38 degrees).

2. Calculate the time of flight:
t = 2 * (27.1 m/s * sin(38 degrees) / 9.8 m/s^2).

3. Calculate the vertical displacement:
h = (27.1 m/s * sin(38 degrees))^2 / (2 * 9.8 m/s^2).

After performing the calculations, the time of flight is approximately 2.3 seconds, and the vertical displacement is approximately 7.5 meters.

Perhaps Alexa can help you.