# Projectile Motion (Firefighter)

1. Nov 16, 2009

### Chandasouk

1. The problem statement, all variables and given/known data

A firefighter d=50 m away from a burning building directs a stream of water from a fire hose at an angle of θ=30° above the horizontal. The speed of the water stream is 40 m/s.

a)
What is the maximum height in the air that the water reaches?

b)
At what height h will the water strike the building?

I found the Velocity for the horizontal component

Vx = cos30*40=34.64m/s

and the vertical component Vy= sin30*40 = 20m/s

For part a, is the height 10.2m?

I did

$$\Delta$$X=Vx*t

50m=34.64m/s*t

t=1.44 seconds

then plugged that in the equation

$$\Delta$$y=Vy*T+1/2at^2

which at max height is $$\Delta$$y=1/2at^2

1/2(-9.8)(1.44)^2 = 10.2m

Or would that be the height of the building?

2. Nov 16, 2009

### Chandasouk

Because it strikes the building at it's highest point?

3. Nov 16, 2009

### rl.bhat

Sorry. I am rechecking your calculations. Your calculation of maximum height is wrong.
First of all find the time taken by water to reach the maximum height where Vy is equal to zero. Your calculation of time taken by the water to strike the building is correct. And Δy is also correct.

4. Nov 16, 2009

### Chandasouk

the $$\Delta$$y=10.2m corresponds to the max height of the water or building?

5. Nov 16, 2009

### rl.bhat

10.2 m is height of the water at 1.44 s. It happens to be the height of the building.
Maximum height is still more.

6. Nov 16, 2009

### Chandasouk

okay, using the equation

Vf=Vi+at

0 = 34.64m/s + -9.8m/s^2t

t = 3.53 seconds

Y = 1/2at^2

Y=1/2(-9.8m/s^2)(3.53 sec) = 61m

So that is the water's maximum height it reaches?

7. Nov 16, 2009

### rl.bhat

Here Vi is Vsinθ.
To get the maximum height you can use
0 = (Vsinθ)^2 - 2*g*H

8. Nov 16, 2009

### Chandasouk

So you would use the Y component of the waters velocity to find the maximum height it reaches?

Vi = sin30*40 = 20m/s

0 = (20)^2-2gh

H = 20.4m

9. Nov 16, 2009

### rl.bhat

That is right.