Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 27.0m/s as it leaves the hose nozzle. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation of the hose until the water takes 3.00s to reach a building 35.0m away. You can ignore air resistance; assume that the end of the hose is at ground level.

Find the speed of the water at the highest point in its trajectory.

How high above the ground does the water strike the building?

How fast is it moving just before it hits the building?

The 35 m of horizontal travel in 3.0 s tell you the horizontal velocity component, 11.67 m/s. The hose angle is the angle whose cosing is 11.67/27.0

At the highest point of the trajectory, the water's vertical velcoity component is zero. You should know what the horizintal component is.. it does not change.

Try to do the rest of this yourself. It is very smiliar to the other questions you have posted this evening. You learn more by doing than by having work done for you.

i know the aswer

To solve this problem, we can break it down into two parts: the horizontal motion and the vertical motion.

1. Horizontal Motion:
Since there is no air resistance, the horizontal velocity remains constant throughout the entire motion. We know the distance covered horizontally is 35.0m, and the time taken is 3.00s. Therefore, the horizontal velocity can be calculated using the formula:
horizontal velocity = horizontal distance / time
horizontal velocity = 35.0m / 3.00s
horizontal velocity = 11.7m/s

2. Vertical Motion:
a. Finding the speed of water at the highest point:
At the highest point of its trajectory, the vertical velocity of the water is zero. We can find the time taken to reach the highest point using the concept of symmetry in projectile motion. Since the time taken to reach the building is 3.00s, the time taken to reach the highest point is half that time, which is 1.50s. Using the equation of motion, we can find the vertical velocity at the highest point:
final vertical velocity = initial vertical velocity + (acceleration * time)
0 = initial vertical velocity - (9.8m/s^2 * 1.50s)
initial vertical velocity = 14.7m/s

b. Finding the height above the ground where the water strikes the building:
To find the height, we need to find the vertical displacement covered by the water. We can use the kinematic equation:
vertical displacement = initial vertical velocity * time + (1/2) * acceleration * time^2
vertical displacement = 14.7m/s * 3.00s + (1/2) * (-9.8m/s^2) * (3.00s)^2
vertical displacement = 22.1m

c. Finding the speed of the water just before it hits the building:
To find the speed just before hitting the building, we can use the final vertical velocity. There is no acceleration in the vertical direction since there is no change in velocity:
final vertical velocity = initial vertical velocity + (acceleration * time)
final vertical velocity = 14.7m/s + (-9.8m/s^2 * 3.00s)
final vertical velocity = -14.9m/s (Negative because the water is moving downward)
speed = √(horizontal velocity^2 + vertical velocity^2)
speed = √(11.7m/s)^2 + (-14.9m/s)^2
speed = √(137.29m^2/s^2 + 222.01m^2/s^2)
speed = √(359.30m^2/s^2)
speed ≈ 18.95m/s

In summary:
a. The speed of the water at the highest point in its trajectory is 14.7m/s.
b. The water strikes the building at a height of 22.1m above the ground.
c. The water is moving downward at a speed of approximately 18.95m/s just before it hits the building.

To solve this problem, we can use the principles of projectile motion. Projectile motion is the motion of an object thrown into the air with an initial horizontal velocity and an initial vertical velocity due to gravity.

1. Finding the speed of the water at the highest point in its trajectory:
First, we need to determine the initial vertical velocity of the water. Since we are given the total time of flight (3.00s) and the distance traveled horizontally (35.0m), we can use the equation of motion:

Distance = Initial Velocity * Time + (1/2) * Acceleration * Time^2

In the vertical direction, the initial velocity is zero at the highest point, and the acceleration is due to gravity (-9.8 m/s^2). The distance traveled vertically is the height of the trajectory.

Horizontal Distance = Initial Horizontal Velocity * Time
35.0m = Initial Horizontal Velocity * 3.00s

Solving for the initial horizontal velocity will give us the initial velocity of the water:

Initial Horizontal Velocity = 35.0m / 3.00s = 11.67 m/s

The speed of the water at the highest point in its trajectory is equal to the magnitude of the initial velocity, so the speed is 11.67 m/s.

2. Finding the height above the ground that the water strikes the building:
To determine the height, we need to find the vertical distance traveled by the water when it reaches the building.

Vertical Distance = Initial Vertical Velocity * Time + (1/2) * Acceleration * Time^2

Again, the initial vertical velocity is zero, and the acceleration is due to gravity (-9.8 m/s^2). We have the total time of flight (3.00s). Solving for the vertical distance will give us the height above the ground the water strikes the building.

Vertical Distance = 0 + (1/2) * (-9.8m/s^2) * (3.00s)^2
Vertical Distance = -44.1m

Since the water is projected upwards, the negative sign indicates that the water is below the starting point. Therefore, the water strikes the building at a height of 44.1 meters above the ground.

3. Finding the speed of the water just before it hits the building:
To find the speed just before the water hits the building, we can use the vertical component of the velocity.

Vertical Velocity = Initial Vertical Velocity + Acceleration * Time

Again, the initial vertical velocity is zero, and the acceleration is due to gravity (-9.8 m/s^2). We have the total time of flight (3.00s). Solving for the vertical velocity will give us the speed just before the water hits the building.

Vertical Velocity = 0 + (-9.8m/s^2) * 3.00s
Vertical Velocity = -29.4 m/s

The speed of the water just before it hits the building is equal to the magnitude of the vertical velocity, so the speed is 29.4 m/s.