Projectile Motion on Inclined Surface Question

In summary, to determine the range of water sprayed at an angle of 90° from a slope at 20m/s, the kinematic equations were utilized. The initial conditions were set at the point of the spout of water and the angle of the plane was found to be 36.869°. The equations for position and velocity were derived, and the parametric equations were used to find the intersection of the two plane curves. A sign error was corrected in the equation for x(t), resulting in an accurate solution for the range of 76.33 meters.
  • #1
ltkach2015
37
1

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.
 

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  • #2
ltkach2015 said:
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.
ltkach2015 said:
6) Find the Intersection of the two plane curves.
You have all you need now, just do it?
ltkach2015 said:
Range = sqrt( xi^2 + yi^2)
Right.

If you introduce a coordinate system, you should briefly explain its orientation.
 
  • #3
ltkach2015 said:
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
mfb said:
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
TSny said:
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, don't know why I included it here)
- the 'i' does mean intersection. apologies.
- I will look into the sign error, hopefully that fixes it.
 
  • #6
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.
 

What is projectile motion on an inclined surface?

Projectile motion on an inclined surface refers to the motion of an object that is launched or thrown on a surface that is not horizontal, but instead has a slope or incline.

What factors affect the trajectory of a projectile on an inclined surface?

The trajectory of a projectile on an inclined surface is affected by the initial velocity of the object, the angle of the slope, and the force of gravity acting on the object.

How does the angle of the slope affect the range of a projectile on an inclined surface?

The angle of the slope affects the range of a projectile on an inclined surface because it determines the horizontal component of the initial velocity. A steeper slope will result in a larger horizontal velocity, leading to a longer range.

What is the difference between projectile motion on a horizontal surface and an inclined surface?

The main difference between projectile motion on a horizontal surface and an inclined surface is that on an inclined surface, the force of gravity acts not only in the vertical direction, but also in the direction of the slope. This affects the trajectory and range of the projectile.

How can you calculate the trajectory and range of a projectile on an inclined surface?

To calculate the trajectory and range of a projectile on an inclined surface, you can use the equations of motion, taking into account the angle of the slope and the force of gravity. You can also use vector components to break down the initial velocity into horizontal and vertical components and analyze the motion separately.

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