Projectile Motion Question - Can someone verify that the question is correct?

[cerealkiller]

Hi guys,

Doing a projectile motion question, I think I know how to do it but I can't get the answer it wants me to.

Anyways:

A tall building stands on level ground.The nozzle of a water sprinkler is positioned at a point P on the ground at a distance d from a wall of the building. Water sprays from the nozzle with speed V and the nozzle can be pointed in any direction from P.

(a) If V > (gd)^(1/2), prove that the water can reach the wall above ground level.

(b) Suppose that V = 2(gd)^(1/2). Show that the portion of the wall that cn be sprayed with water is a parabolic segment of height 15d/8 and area (5/2)*(d^2)*(15)^(1/2).

Sorry about my horrible typesetting, I'm not familiar with latex.

I can get (a) out, and given the height of 15d/8 I can get the right expression for area, but no matter how I do it I end up with the height of the parabolic segment being 3d/4.

I know it would be a huge pain to type out the solutions, but could someone give it a go and see if they can get the height of the parabolic segment as 15d/8? I've checked over my work a few times and end up with 3d/4.

Thanks

 

[mfb]

How did you calculate the area?
The volume which can be reached should be a paraboloid. It has two (relevant) parameters, which can be fixed via the maximal height above the nozzle (vertical water flow) and the maximal height at the building.
The intersection between the paraboloid and your building then gives a parabola.

If that is different from your approach, try this to check your result :).

 

[cerealkiller]

If you assume the parabola has height 15d/8 you can pretty easily figure out how wide the parabola will be (I got it to be 2d√15), then using Simpson's rule you can get the correct expression for area, I just can't get to the 15d/8 part.

Not sure what I'm doing wrong, I let x = d, then find an expression for time, then substitute this value for time into the equation for vertical displacement, and you should end up with the maximum height of the parabola right?

 

[azizlwl]

(b) your can't solve without the angle.
Or is the the same (45°) as in question (a)

 

[cerealkiller]

Ok, that was my mistake. I assumed that because the max range of the spray would occur at 45° then the maximum height would for a given velocity at a given horizontal range would occur at the same angle.

Forming an equation with y and θ when x=d, where θ is the initial projection velocity, then finding the max. of that equation gives y=15d/8.

So the max height on the wall is given by a different value of θ, depending on the direction (not just vertically) that the nozzle is pointed.

This question is deceptively complicated.

 

[cerealkiller]

Ok, I've done all of it except actually prove that the path is a parabolic segment. All of my algebra bashing comes down to proving that something of the form

y = x^2 + (d^2 +x^2)^(1/2)

is a parabola (where d is a constant). Is that trivial or am I missing something? I've used some graphing tools to graph y and it definitely looks like a parabola. Is it fair to assume that something in the form of ^^^ is a parabola?

 

[Infinitum]

Ok, I've done all of it except actually prove that the path is a parabolic segment. All of my algebra bashing comes down to proving that something of the form

y = x^2 + (d^2 +x^2)^(1/2)

is a parabola (where d is a constant). Is that trivial or am I missing something? I've used some graphing tools to graph y and it definitely looks like a parabola. Is it fair to assume that something in the form of ^^^ is a parabola?

Assuming your equation is correct, I can help you to prove whether it is a parabola

So you have,

[tex]y = x^2 + \sqrt{(d^2 +x^2)}[/tex]

Here, the simplification of [itex]\sqrt{(d^2 +x^2)}[/itex] will yield a linear polynomial in x, of the form [itex]px+q[/itex], where p and q are constants. So your equation becomes,

[tex]y = x^2 + px + q[/tex]

Can you take it from here and prove it is of the form [itex]X^2 = 4aY[/itex]?