Ask Experts Questions for FREE Help !
Ask
    SimonJensen's Avatar
    SimonJensen Posts: 1, Reputation: 1
    New Member
     
    #1

    Sep 11, 2017, 12:45 PM
    Projectile Motion On An Inclined Plane
    I was given the following information:
    A ball is thrown with an initial velocity V0 atan angle of φ at an inclined planethat has an angle of θ with the horizontalline.

    I really need and would appreciate some help to work of the following:
    * The x-coordinate for the point where the ball touches the inclined plane (expressed by V0, g, θ and φ.

    I have attached a link to an illustration I made of the problem.

    https://ibb.co/hNKk6v

    Thank you!
    ThomasCole's Avatar
    ThomasCole Posts: 1, Reputation: 1
    New Member
     
    #2

    Sep 12, 2017, 06:07 AM
    Want to know the answer too.
    ma0641's Avatar
    ma0641 Posts: 15,681, Reputation: 1012
    Uber Member
     
    #3

    Sep 12, 2017, 10:21 AM
    Must have the same homework.
    smoothy's Avatar
    smoothy Posts: 25,495, Reputation: 2853
    Uber Member
     
    #4

    Sep 12, 2017, 08:57 PM
    Both of you should consider vocational Tech curriculum if you can't do the homework in the one you are in.

    The site rules REQUIRE you to show us your work and your answer, something neither of you have done.

    Its a VIOLATION of site rules to hand out homework answers so you don't have to do your own work.
    MBarosM's Avatar
    MBarosM Posts: 5, Reputation: 1
    New Member
     
    #5

    Sep 13, 2017, 09:18 AM
    I wasn't aware of the site's rules, so I apologize, as I was genuinely looking for help. Here is my own attempt to solve it, can anyone tell me if I am on the right track?:

    My method was solving this equation:
    0 = (tan(θ) - tan(φ)) - x( g / (2 * cos^2 (θ) * V0^2)

    It has two solutions, either x = 0 or (tan(θ) - tan(φ)) - x( g / (2 * cos^2 (θ) * V0^2) = 0

    Then I tried to solve the equation for x, and got the following as my result:
    x = ((tan(θ)-tan(φ)) * (2 * V0^2 * Cos^2(theta)) / g when I try
    ebaines's Avatar
    ebaines Posts: 12,130, Reputation: 1307
    Expert
     
    #6

    Sep 13, 2017, 10:46 AM
    Quote Originally Posted by MBarosM
    My method was solving this equation:
    0 = (tan(θ) - tan(φ)) - x( g / (2 * cos^2 (θ) * V0^2)
    I get something a bit different:



    It seems you have the definitions of phi and theta reversed. Also, you didn't specify this but I assume that the launch point is at the base of the incline, correct?
    MBarosM's Avatar
    MBarosM Posts: 5, Reputation: 1
    New Member
     
    #7

    Sep 13, 2017, 11:40 AM
    Illustration of the situation: https://ibb.co/fSvwEF

    Calculations:https://ibb.co/dtDBgv

    If you are not able to read my handwriting, I did the following:

    Y-direction: y = V0 * Sin(Theta) * t * 1/2 * g * t^2

    X-direction: x = V0 * Cos(Theta) * t

    I den solved for t in the equation for x and got:t = x / (V0 * Cos(theta)

    I inserted the above value for t in the equation for y:

    y = V0 * sin(theta) * (x / (V0 * cos(theta)) - g/2 *( x / (V0 * Cos^2(theta))

    I substituted the left side of the equation (y) with tan(phi) * x and got:
    tan(phi) * x = x * tan(theta) - g/2 * ( x / (V0 * Cos^2(theta))

    I got the left side of the equation on the right side and got:
    0 = x * (tan(theta) - tan(phi)) - x * ( g / (2 * V0^2 * Cos^2(theta))

    After this, I solve the equation with respect to x (it is not on the attached picture), but I have done it:

    0 = x * (tan(theta) - tan(phi) - x * ( g / (2 * V0^2 * Cos^2(theta))

    x * ( g / (2 * V0^2 * Cos(theta))^2 = (tan(theta) - tan(phi))

    x = ((tan(theta) - tan(phi)) * 2 * V0^2 * Cos^2(theta)) / g
    ebaines's Avatar
    ebaines Posts: 12,130, Reputation: 1307
    Expert
     
    #8

    Sep 13, 2017, 01:48 PM
    Quote Originally Posted by MBarosM
    Y-direction: y = V0 * Sin(Theta) * t * 1/2 * g * t^2

    X-direction: x = V0 * Cos(Theta) * t
    Two issues with this. First, the second half of that first equation should be "- 1/2 g t^2", not "* 1/2 g t ^2." This appears to be a simple typo, based on subsequent equations, so no harm done. The other issue is more critical: the equation of motion for the projectile is dependent on the angle of launch as measured from the horizontal. Where you use theta it should be theta + phi, because the sum of those two angles is the angle of the initial velocity V0 relative to horizontal. Please note that I made a similar mistake in my previous post - I hadn't had a chance to see the figure that was included with the original post, so assumed (incorrectly) that the angle phi was measured relative to horizontal. But now I see that phi is measured relative to the inclined plane, and hence the correct equation for Y-direction motion is:

    y = V0 sin(theta + phi) t - (1/2) g t^2.

    And for x-direction we have:

    x = V0 cos(theta + phi) t.

    We're interested in the point in time when the projectile hots the ramp, at which point y/x = tan(theta), so you can replace y in the first equation with x tan(theta). Can you take it from here?
    MBarosM's Avatar
    MBarosM Posts: 5, Reputation: 1
    New Member
     
    #9

    Sep 13, 2017, 02:05 PM
    After I substitute tan(theta) * x for y in the equation, I have:

    tan(theta) * x = tan(theta + phi) * x - g/2 * ( x^2 / (V0^2 * Cos^2 (theta + phi)

    0 = tan(theta + phi) * x - tan(theta) * x - g/2 * ( x^2 / (V0^2 * Cos^2 (theta + phi)

    0 = ( (-g * x^2) / (2 *cos^2(theta + phi) * V0^2)) + tan(theta + phi) * x^2 - tan(theta) * x

    Am I on the right track now?
    ebaines's Avatar
    ebaines Posts: 12,130, Reputation: 1307
    Expert
     
    #10

    Sep 14, 2017, 05:46 AM
    Looking good. Now divide through by x and rearrange to get x by itself.

Not your question? Ask your question View similar questions

 

Question Tools Search this Question
Search this Question:

Advanced Search

Add your answer here.


Check out some similar questions!

Ball in an inclined plane [ 2 Answers ]

Help me to understand the my physics homework Suppose if a ball is rolling down in an inclined plane, what happens to the normal force acting on it? How to understand the Normal force in this situation Please explain Thanks

Ta ball is released from rest and rolls down an inclined plane, requiring 4s to cover [ 1 Answers ]

How many cm would it have fallen vertically in the same time?

A body slding on a smooth inclined plane takes 4sec to reach the bottom, starting fro [ 0 Answers ]

A body slding on a smooth inclined plane takes 4sec to reach the bottom, starting from rest at the top. How much time does it take to cover 1/4th dist from the top?

When you are given an inclined plane must the degree be given ? [ 1 Answers ]

Must the angle from the ground to the inclined plane be given to you as part of the question ?

Inclined plane [ 5 Answers ]

I need to know how to calculate force transferred from the perpendicular to horizontal assuming zero friction. Example: 10 pounds of verticle force applied to a 30 degree plane will move the plane horizontally with how much force? Or to put it another way, how much force is lost in the example, and...


View more questions Search