# Homework Help: Horizontal projection

1. Sep 20, 2008

### tater08

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

A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0 degrees above the horizontal. The firefighters want to direct the water at a blaze that is 10.0 m above ground level.

How far from the building should they position their cannon? There are two possibilities (d_1<d_2); can you get them both? (Hint: Start with a sketch showing the trajectory of the water.)

2. Relevant equations

x=vix*t
vfx=vix
vfx^2=vix^2
y=viy*t+0.5a8t^2
Vfy=viy+a*t
vfy^2=viy^2+2*a*y
3. The attempt at a solution

i do not know where to start with this problem. Please and Thank you for the help!!

2. Sep 20, 2008

### LowlyPion

What is the equation for the motion of the water through the air in the y direction and the x direction?

At what times will the height of the water satisfy the y position equation?

What does that mean in terms of how far away you need to be using your x direction equation?

3. Sep 21, 2008

### tater08

Knowns:
y component

Yo=0
y=10
vyo=25sin53
=19.97
a=9.8

X component

xo=?
xf=0
vxo=25sin53
t=
a=0

y=yo+voyt-0.5at^2 10=25sin53t-0.5 9.8 t^2 it can form the quadratic equation so -4.9t^2+25sin53t-10 t=0.58 and 3.49

and i take those numbers and enter them into 25cos53 t =x i get 8.7 and 51.3 but the answer that has 3.49 seconds is wrong. Does anybody know what i am doing wrong?

4. Sep 21, 2008

### LowlyPion

Using .6018 as Cos 53, I get 15.04 as horizontal velocity and the 3.49 yields 52.5 m. And 8.8 for the closer one.

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