Finding Angle: Horizontal Water Projection at 12m - Help Needed

[astrololo]

Homework Statement

A fountain projects water horizontaly at a height of 12 m. The water goes on the floor after having traveled 15 m horizontally. Calculate the angle the water does with the floor.

Homework Equations

I know that the accelerate is constant, so its one of the cinematic equation. Not sure of which one.

The Attempt at a Solution

I have no idea on how to do this. I know that I must decompose my situation into two components, one for the x's and the other for the y's. Any hint would be helpful !

 

[gneill]

You could approach the problem via kinematic equations, but that may be the long way around.

Consider what type of curve a projectile describes. If you can find an equation to fit your given data then you should be able to find the slope (hence the angle) at any point along the curve...

 

[astrololo]

You could approach the problem via kinematic equations, but that may be the long way around.

Consider what type of curve a projectile describes. If you can find an equation to fit your given data then you should be able to find the slope (hence the angle) at any point along the curve...

Vx^2 = Vx initial^2 + 2a(x-xinitial) ? I know that its a parabola

 

[astrololo]

Help please !

 

[gneill]

Okay, ignoring the kinematic view for the moment, and thinking in terms of functions from math class, let's start with a general expression for a parabola that would suit the given situation:

y(x) = ?

 

[astrololo]

y=-x^2+c

 

[gneill]

y=-x^2+c

Close. You need to add a coefficient to the $x^2$ term to account for the "width" of the parabola. The "c" term takes care of the vertical offset. So write:
$$y = -a x^2 + c$$
Can you find a and c using the known data points?

 

[astrololo]

Close. You need to add a coefficient to the $x^2$ term to account for the "width" of the parabola. The "c" term takes care of the vertical offset. So write:
$$y = -a x^2 + c$$
Can you find a and c using the known data points?

12 and 15

 

[gneill]

12 and 15

Justify.

 

[astrololo]

Justify.

Oh I actually have (0,12) : (15,0) Which I can use I think. to find the general equation.

 

[astrololo]

ok i found y=-4/75x^2 + 12

 

[gneill]

ok i found y=-4/75x^2 + 12

Excellent. Now can you find the angle where it meets the "floor"? Think: slope, tangent,...

 

[astrololo]

Excellent. Now can you find the angle where it meets the "floor"? Think: slope, tangent,...

Are you suggesting that Ishould use differential calculus ?

 

[gneill]

Are you suggesting that Ishould use differential calculus ?

Unless you can dig up a "canned formula" for the slope of a parabola from your Functions class notes, I'd suggest calculus, yes.

 

[astrololo]

Unless you can dig up a "canned formula" for the slope of a parabola from your Functions class notes, I'd suggest calculus, yes.

Then this isn't the right way to do it. I know calculus, but I'm not supposed to use it now.

 

[gneill]

Then this isn't the right way to do it. I know calculus, but I'm not supposed to use it now.

Ah. That's a shame, because the work is essentially done

So I guess it's back to kinematics then. I suggest you start by finding the vertical velocity at impact, and the time to impact. You can do this because you know that the vertical and horizontal components can be treated separately, and you know the distance and acceleration involved.