Projectile Motion on Inclined Surface Question

  • Thread starter ltkach2015
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  • #1
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Homework Statement



Water is sprayed at an angle of 90° from the slope at 20m/s. Determine the range R.

PLEASE SEE ATTACHMENT


Homework Equations


[/B]
Kinematic Equations:
acceleration: A = 0i +g(-j) + 0k;
velocity: dV/dt = A;
position: dR/dt = V;

Origin set at the point spout of water.

Angle of the plane: θ = atand(3/4) = 36.869°

Magnitude of acceleration: g = 9.81;

Initial Conditions
Velocity: Vb(t=0) = 10; => Vb(t=0) = Vb*cosd(90-θ)*(-i) + Vb*sind(90-θ*(-j) + 0*(-k)
Position: Rb(t=0) = 0*(i) + 0*(j) + 0*(k)


The Attempt at a Solution




1) integrate acceleration: A = [0; -g ;0] = dVb/dt

∫dVb = ∫ [0; -g; 0] dt

=> Vb(t) = c1*(i) + (-g*t + c2)*(-j) + c3*(k);

impose initial conditions: Vb(t=0) = Vb*cosd(90-θ)*(-i) + Vb*sind(90-θ)*(-j) + 0*(-k)
=> c1 = Vb*cosd(90-θ); c2 = Vb*sind(90-θ); c3 = 0;

2) integrate velocity:

dR = Vb(t)dt = Vb(t) = [-Vb*cosd(90-θ); g*t + Vb*sind(90-θ); 0]dt;

=> R(t) = (Vb*cosd(90-θ)*t + c4)*(i) + (-g/2*t^2 + Vb*sind(90-θ)*t + c5)*(j) + c6*(k)

impose initial conditions: Rb(t=0) = 0*(i) + 0*(j) + 0*(k)
=> c4 = 0; c5 = 0; c6 = 0;

3) List of Derived Equations:

R(t) = (-Vb*cosd(90-θ)*t )*(i) + (-g/2*t^2 + Vb*sind(90-θ)*t )*(j) + 0*(k)

Vb(t) = (-Vb*cosd(90-θ))*(i) + (-g*t + Vb*sind(90-θ))*(j) + 0*(k)

4) Equation of Inclined Plane:
f(x) = 3/4*x

5) Parametrize the position function:

x(t) = (-Vb*cosd(90-θ)*t )
=> t = x(t)/(Vb*cosd(90-θ)

y(x(t)) = Vbsind(90-θ)*{x(t)/(Vb*cosd(90-θ))} - g/2*({x(t)/(Vb*cosd(90-θ))}^2

6) Find the Intersection of the two plane curves.

7) Use magnitude of x and y to find Range: (probably didn't say that correctly)

Range = sqrt( xi^2 + yi^2)


I would appreciate all the help I can get. Thank you.
 

Attachments

  • Screen Shot 2015-08-14 at 1.10.52 PM.png
    Screen Shot 2015-08-14 at 1.10.52 PM.png
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Answers and Replies

  • #2
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R(t) = (-Vb*cosd(90-θ)*t )*(i) + (-g/2*t^2 + Vb*sind(90-θ)*t )*(j) + 0*(k)
I guess you could have started here.
6) Find the Intersection of the two plane curves.
You have all you need now, just do it?
Range = sqrt( xi^2 + yi^2)
Right.

If you introduce a coordinate system, you should briefly explain its orientation.
 
  • #3
TSny
Homework Helper
Gold Member
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5) Parametrize the position function:
x(t) = (-Vb*cosd(90-θ)*t )
=> t = x(t)/(Vb*cosd(90-θ)

Looks like a sign error solving for t. Your equation for x(t) has a negative sign that disappears when solving for t.

In general your work looks good, although you have written it out in a very formal manner. Nothing wrong with that, but it is a bit lengthy.

Just curious: when you write cosd for the cosine function, does the "d" denote that the argument is in degrees?

Range = sqrt( xi^2 + yi^2)
OK. But what does the "i" stand for in xi and yi? "Intersection"?

I would appreciate all the help I can get. Thank you.

Not sure what help you need. It all looks good except for the sign error mentioned above.
 
  • #4
37
1
I guess you could have started here.
You have all you need now, just do it?
Right.

If you introduce a coordinate system, you should briefly explain its orientation.



Hi mfb, thank you for the quick response.
- Oh! Yes my my coordinate system is a right handed one: x positive right, y positive up
- Unfortunately, my final answer is different than the solution.

ANS: Range = 76.5 meters

Range = sqrt( xi^2 + yi^2) = sqrt( 17.126^2 + 12.8449^2) = 21.4 meters


-wait that doesn't make any sense these should be negative values...
 
Last edited:
  • #5
37
1
Looks like a sign error solving for t. Your equation for x(t) has a negative sign that disappears when solving for t.

In general your work looks good, although you have written it out in a very formal manner. Nothing wrong with that, but it is a bit lengthy.

Just curious: when you write cosd for the cosine function, does the "d" denote that the argument is in degrees?


OK. But what does the "i" stand for in xi and yi? "Intersection"?



Not sure what help you need. It all looks good except for the sign error mentioned above.



Hi TSny, thanks for the quick reply.

- the 'd' in the cosd does denote that the argument is in degrees (from matlab, dunno why I included it here)
- the 'i' does mean intersection. apologies.
- I will look into the sign error, hopefully that fixes it.
 
  • #6
37
1
Hey TSny!

Thank you it was a sign error!

xi = -61.07
yi = -45.80

Range = sqrt[ (-61.07)^2 + (-45.80)^2 ] = sqrt(5827.17) = 76.33. close enough

Thank you both for all your help. I was really confused.
 

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