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Kinematics Equations and a Cliff

  1. Dec 11, 2014 #1
    1. The problem statement, all variables and given/known data

    A car drives off a cliff that is 100m high. It has to land in water and the water starts 30m away from the cliff. Its goal is to land 90m into the water. How fast must the car be going to land at that point in the water. Air resistance is negligible.

    v=0
    v0=?
    a= -9.91m/s2
    Δy=100m
    Δx=120m

    2. Relevant equations

    My plan was to use kinematics and determine the time it takes for the car to fall using Δy= vt - 1/2(a)t2, then put that into Δx= 1/2(v+v0)t. Isn't it true that whether you throw something horizontally or just drop it, it takes the same time to hit the ground? Why does that not apply here regarding the horizontal distance and velocity?

    3. The attempt at a solution


    My answer is wrong. I get 53 m/s (this seems like a lot anyway).


     
  2. jcsd
  3. Dec 11, 2014 #2

    Bystander

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    Think over the last part of your strategy, keeping in mind a condition from the initial problem statement.
     
  4. Dec 11, 2014 #3

    Delphi51

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    The strategy is certainly correct. I get a much smaller answer, though.
    I find your horizontal motion distance formula confusing with the two velocities and division by 2.
    The horizontal velocity is constant; why not just call it v? Try running it through that way!
     
  5. Dec 11, 2014 #4

    ehild

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    a=-9.81 m/s2
    That means you took upward positive. Will the car move upward when falling from the cliff????? What is the sign of Δy?

    Δy= vt - 1/2(a)t2 is true if v means the vertical component of the initial velocity. But the car drives on a horizontal cliff. What is the vertical component of velocity at the instant when it leaves the cliff??
    Δx= 1/2(v+v0)t is valid for uniform accelerating motion in the x direction, (horizontally). Is there any horizontal force applied on the car, causing acceleration in the horizontal direction, and changing the horizontal component of velocity?And you can not use the same notation v for the initial vertical velocity and the final horizontal velocity components.

    It is true that whether you throw something horizontally or just drop it, it takes the same time to hit the ground.
     
  6. Dec 11, 2014 #5
    First of all whether you throw something horizontally or dropping it does take the same time. However this does not mean it goes the same horizontal distance, the x-component of the velocity determines that. Also there is no velocity in the y-direction as the car is moving horizontally. What you are going to have to do is find the time it takes the car to hit the ground using
    y = y0-1/2*g*t^2, Where g = 9.8 m/s^2 and y0 is the initial height.
    Then use ∆x = v0t. Use the time (t) you solved for in the y equation and plug it into this x equation to solve for v0.
     
    Last edited: Dec 11, 2014
  7. Dec 11, 2014 #6
    Air resistance being negligible is part of the problem and becomes in incorrect assumption for part C. It's there to show us that the final answer, whatever it is, will be inaccurate... I am having trouble getting to that final answer.
     
  8. Dec 11, 2014 #7

    Delphi51

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    I don't see any parts in the question posted. What is part C?
     
  9. Dec 11, 2014 #8

    ehild

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    @velo city: The policy of the forum does not allow to solve the problem for the OP. And it is even worse to show wrong solution, as yours is for the time for the fall. Also your formula for the ∆x is misleading as the OP used the notation a for -9.81.
     
  10. Dec 11, 2014 #9
    I thought that v was v final which is 0 with regards to the vertical portion of its fall? I thought v0 was the initial velocity which is what I am trying to find.
     
  11. Dec 11, 2014 #10
    I did not post that part because I understood it.
    I was just letting you know, so people don't try to factor in air resistance mathematically.
     
  12. Dec 11, 2014 #11
    sorry i will change it.
     
  13. Dec 11, 2014 #12

    Delphi51

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    The v and vo are in the horizontal motion formula with no acceleration. We used to write d = vt in high school. This v is the speed of the car you are trying to find.
     
  14. Dec 11, 2014 #13

    ehild

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    The vertical and horizontal motions of the car are independent if air resistance is ignored. The vertical motion is free fall. The displacement is Δy= vy0t - 1/2(g)t2
    where vy0 is the initial vertical velocity. What is its value?
    The velocity of the horizontal motion does not change during the fall. It is the unknown v.
     
  15. Dec 11, 2014 #14
    By the way I was using the time for the fall to find the x displacement. So my solution was not wrong.
     
  16. Dec 11, 2014 #15
    I don't think the car has an initial vertical velocity right?
     
  17. Dec 11, 2014 #16
    0?

    Anyway, I have the solution in front of me. It has a bunch of trig which I do not understand. The solution used the "standard coordinate system with the origin at the initial position of the car."

    x = x0 + (v0,x)t+(1/2)(ax)t2 = 0 + v0(cosθ)t+(1/2)(0)t2=v0(cosθ)t
     
    Last edited: Dec 11, 2014
  18. Dec 11, 2014 #17
    the trig comes only from projectiles launched at an angle. In this case that angle (ø) is 0 because the car is only moving horizontally.
     
  19. Dec 11, 2014 #18
    You should probably watch this video: . It will help you understand if you watch the whole thing. I don't want to give you a solution cause that is against the rules. So that video should help.

    Also watch this one:
     
  20. Dec 11, 2014 #19

    ehild

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    The time (more thn 20 s) was wrong. (You removed it already ). Also, the notation was misleading in the formula of x displacement. What value for a did you use? As the OP set a=-9.81 m/s2.
     
  21. Dec 11, 2014 #20
    I used a =0 as there is no acceleration in the x direction
     
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