Projectile Motion: stream of water

[kat R]

Homework Statement

(For this problem ignore the height of the firefighter) A firefighter, a distance (d) from a building, shoots a stream of water at an initial angle, above the horizontal, at initial speed (Vo). At what height (h) does the water strike the building?

No values were given to me, so I have no idea how to solve for h.

Homework Equations

Yf=Yi+Vyi*t+.5*ay*t^2

There could be a better equation for this, but I started here.

The Attempt at a Solution

I'm not entirely sure if I am supposed to actually find a way to solve for h or if I just need to set up a formula so it solves for h.

Can anyone help me out? Thanks!

 

[jbriggs444]

Homework Equations

Yf=Yi+Vyi*t+.5*ay*t^2

You wrote down an equation for y, the height of a droplet/spurt that has been flying for t seconds. Can you write down a similar equation for its horizontal position, x?

 

[kat R]

Xf=Xi+Vxi*t+.5*ax*t^2

 

[jbriggs444]

Xf=Xi+Vxi*t+.5*ax*t^2

Let us work on simplifying this equation. Can we get rid of any terms?

[Looking ahead, can we solve for t in terms of any or all of d, $v_0$, the angle $\theta$ and h]

 

[kat R]

I think Xi could be taken out but I don't think anything else can

 

[jbriggs444]

I think Xi could be taken out but I don't think anything else can

You should be able to work out a way to replace Xf and Xi with d.
What about $a_x$? What is its numeric value?

 

[kat R]

Yea Xf-Xi would equal d since Xi is 0. Depends if acceleration due to gravity would matter in the x-direction

 

[jbriggs444]

Yea Xf-Xi would equal d since Xi is 0. Depends if acceleration due to gravity would matter in the x-direction

Does gravity exert any force in the x direction?
Then what is the resulting acceleration in the x direction?

[One assumes that you have chosen x to be horizontal? Have you?]

 

[kat R]

Okay so ax would be 0 which would leave the equation at d=Vxi*t

 

[jbriggs444]

Okay so ax would be 0 which would leave the equation at d=Vxi*t

We can leave $V_{x_i}$ that way for now. Can you solve for t?

 

[kat R]

So if the equation is d=Vo*cos(θ)*t I can say that t=d/(Vo*cos(θ)) but I can't go any further because I wasn't given any numerical values

 

[jbriggs444]

So if the equation is d=Vo*cos(θ)*t I can say that t=d/(Vo*cos(θ)) but I can't go any further because I wasn't given any numerical values

Having no numerical values is fine. It actually makes things easier. Instead of carrying numbers through your algebra, you can use symbols. Easier to write down and less opportunity for error.

Now then we have a formula for t in terms of the givens of the problem (d, $V_0$, $\theta$). Let that sit for a moment.

Let us go back to your first equation for Y. Can you simplify that at all? What about trying to remove $Y_f$ and $Y_i$ and replacing them with h?

When that's done, substitute in the formula for t and see what you have.

 

[kat R]

So I replaced the Yf and Yi for h and replaced Vyi and got h=Vo*sin(θ)*t+.5*ay*t^2
Then the final equation I got was h=Vo*sin(θ)*[d/(Vo*cos(θ)]+.5*ay*[d/Vo*cos(θ)]^2

 

[jbriggs444]

So I replaced the Yf and Yi for h and replaced Vyi and got h=Vo*sin(θ)*t+.5*ay*t^2
Then the final equation I got was h=Vo*sin(θ)*[d/(Vo*cos(θ)]+.5*ay*[d/Vo*cos(θ)]^2

Good.

There is some simplification that can still be done. In the first term you have $V_0$ in both numerator and denominator. And you have $\sin \theta$ in the numerator and $\cos \theta$ in the denominator. You can cancel the one and rewrite the other.

There is also a dangling unknown, $a_y$. Can you rewrite that in terms of g, the acceleration of gravity?

 

[kat R]

So it would be h=d*tanθ+.5*g*(d/Vo*cosθ)^2

 

[jbriggs444]

So it would be h=d*tanθ+.5*g*(d/Vo*cosθ)^2

One minor problem. "g" is +9.8 m/sec^2. Which direction does the acceleration of gravity point?

 

[kat R]

It points in the -y direction... so the final equation would be h=d*tanθ-.5*g*(d/Vo*cosθ)^2

 

[jbriggs444]

It points in the -y direction... so the final equation would be h=d*tanθ-.5*g*(d/Vo*cosθ)^2

Yes indeed.

Now then, I have not carefully checked your algebra. The resulting equation looks good to me. It has the dependencies that I expected. But it would be good to do a sanity check. [And I don't know that your expectations match mine] So let's run through a test case or two.

$\theta = 0$ (horizontal)
$V_0 = 1$ (meter per second)
d = 1 (meter)
h = ??

One would expect the spurt of water to take 1 second to hit the wall and drop by $\frac{g}{2}$ meters. The formula says... $\frac{g}{2}$ down.

$\theta = 90$ (vertical)
$V_0 = 1$ (meter per second)
d = 1 (meter)
h = ??

One would expect the spurt of water to never get there. The formula says... Divide by zero error. Nice!

$\theta = 60$ (steep angle)
$V_0 = 1$ (meter per second)
d = 1 (meter)
h = ??

One would expect the spurt of water to take 2 seconds to get there, moving 2*sin(60) meters upward due to its initial velocity and 2g meters downward due to gravity. Whipping out a calculator, that's 1.72 up and 19.6 down for a total of 17.88 down.

The formula... agrees.

 

[kat R]

That works! Thank you so much!