# Projectile Motion on Inclined Surface Question

1. Aug 14, 2015

### ltkach2015

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

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

2. Relevant equations

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)

3. 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.

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2. Aug 14, 2015

### Staff: Mentor

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.

3. Aug 14, 2015

### TSny

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.

4. Aug 14, 2015

### ltkach2015

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: Aug 14, 2015
5. Aug 14, 2015

### ltkach2015

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. Aug 14, 2015

### ltkach2015

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.