# Projectile motion of a fire hose

In summary, the water from the hose moves in projectile motion and the firemen adjust the angle of elevation to achieve a water strike at 3 m above the ground.
Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed of 25.0 {\rm m/s} as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation \alpha of the hose until the water takes 3.00 {\rm s} to reach a building 45.0 {\rm m} away. You can ignore air resistance; assume that the end of the hose is at ground level.

I found the following:
angle a = 53.1
speed of water at highest point = 15m/s
magnitude of acceleration at highest point = 9.8m/s^2

I need to find:
How high above the ground does the water strike the building?
How fast is it moving just before it hits the building?

I tried using V^2 - Voy^2 = -2g(y-yinitial)

I can't seem to get the right answer.

I figured out how to solve for the height above the ground and got 15.9m which is correct.
Now I need How fast is it moving just before it hits the building?

How fast it was going.

That means they want the horizontal velocity and the vertical velocity at that point.

The horizontal is easy. V*cosθ

The vertical at that height isn't all that hard either, if you recognize that in a gravity field the speed it's going up at, at a certain height, is the same it is coming down at, at the SAME height.

Hence figure how fast using you V2 = (Vo*Sinθ)2 - 2*g*y

Once you have the 2 components of velocity let Pythagoras figure the rest for you.

Thanks! Got it!

Great. Good luck.

I need help getting started on this problem. The book doesn't really explain where to go so please help!

Jan

Welcome to PF.

The angle is not given, but they give you how to get it.

I needs to go 45 m in 3 s. That means that the horizontal velocity = VoCosθ = 45/3 = 15. If VoCosθ = 15 then Cosθ = 15/25

It also then gives you your vertical component of velocity, because that is VoSinθ.

With vertical velocity you can figure the Vy at the window from

Vy = Voy - g*3

where 3 is the time to get there.

With the vertical and horizontal components of velocity known you can determine |Velocity| at the window.

## 1. What is projectile motion?

Projectile motion is the curved path that an object takes when it is launched into the air and only affected by the force of gravity.

## 2. How does a fire hose demonstrate projectile motion?

A fire hose demonstrates projectile motion when it is used to extinguish fires. The water is propelled out of the hose at a high velocity, following a curved path due to the force of gravity.

## 3. What factors affect the projectile motion of a fire hose?

The factors that affect the projectile motion of a fire hose include the initial velocity of the water, the angle at which the hose is held, and any external forces such as wind or obstacles in the path of the water.

## 4. How can the range of a fire hose be maximized?

The range of a fire hose can be maximized by adjusting the angle at which it is held. The optimal angle is typically between 30-45 degrees, which allows for the greatest horizontal distance covered by the water.

## 5. Why is understanding projectile motion of a fire hose important?

Understanding projectile motion of a fire hose is important for firefighters to effectively extinguish fires and protect themselves and others from potential harm. It also allows for precise and efficient use of water, which is a valuable resource in emergency situations.

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