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Using entry level physics to calculate minimum velocity required.

  1. Feb 12, 2013 #1
    1. The problem statement, all variables and given/known data

    A person is trying to jump over the pit in a car. I need to find what is the minimum speed required to make it over. This speed is calculated not from the start of the ramp, but from the take off.

    Drawing of all the elements

    Please excuse my ugly drawing :P.




    2. Relevant equations
    Projectile motion equations (Where Time is a constant)
    d=vot+1/2at^2


    3. The attempt at a solution
    So, I am trying to find the minimum velocity required to make this jump. We know the θ of the Ram is 25°. We also know that the left side is 15m, and the right is 16m. From there, we can judge that the Ramp is more like this.

    Illustration of my triangle

    We know the right side is 1m higher than the left, so we have to account for that in making our triangle.

    We can calculate our 2nd θ using Tan = 1/15. This yields a theta of 3.8°.

    To calculate the velocity required, i used projectile motion (using time as the only constant)

    Vx | Vy
    d= 15m Vf= X
    V=? Vo=0
    T= ? <--------> T=?
    d=1m
    a=-9.8m/s^2


    Calculate T using d=VoT+1/2at^2
    1= 0 + 1/2(9.8)t^2
    1=4.9t^2
    t=0.45s

    Calculate how fast the person needs to go to make the jump

    D=V * T
    15 = 0.45V
    V= 33 m/s


    Does that make sense? I feel like I did something wrong as I did not use the 25°.

    Any help is much appreciated, and if you need more info/clarification, I will obliged.
     
  2. jcsd
  3. Feb 12, 2013 #2

    cepheid

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    My approach to solving this problem would be to say that we want to solve for the initial speed, v0. In projectile motion, since there is no horizontal acceleration, the velocity in the x direction is constant:

    vx = v0cosθ = const.

    Now, for motion at a constant speed, it's just true that distance = speed*time. So the horizontal range Δx of the projectile is just:

    Δx = vxt = (v0cosθ)t

    You know that Δx must be 15 m to span the gap, so you can solve for t in terms of v0 and θ. That's one unknown eliminated (the time spent in the air).

    Now you switch to the equations of motion for the vertical (y) direction. We have position vs. time given by:

    y = y0 + v0yt - (1/2)gt2

    = (v0sinθ)t - (1/2)gt2

    For convenience, choose the initial vertical position (the top of the ramp) y0 to be 0 m. (It may be 15 m above your ground level, but why not measure heights from the top of the ramp?) So, that makes your final vertical position y = 1 m. So you can find the t that satisfies this equation for y = 1 m, and set that equal to the t from before.
     
  4. Feb 13, 2013 #3
    How would I solve X=Vxt? I only know X=15, which leaves 2 variables.

    I believe I did it backwards from you, finding the time required in the Y, and
    using it in my Vx. Does that make my answer false?

    Also, can the Theta not be found using just Tan=1/15?.
     
  5. Feb 13, 2013 #4

    cepheid

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    Read my post more closely. vx = v0cosθ, so this equation allows you to solve for t in terms of v0, which is the thing you're trying to solve for, and θ, which is given, NOT unknown (see below)

    The order in which you use the equations doesn't matter. The y equation will give you the time required to reach a height of 1 m at a given launch speed, and then you can plug that into your x equation to solve for the launch speed that gives you the right horizontal range.

    That's the wrong theta. You want the launch angle of the projectile, which is the ramp angle, which is already given to be 25°. So, as I said above, θ is already known.
     
  6. Feb 14, 2013 #5
    So, solving for T in terms of Vo would give you:

    t= 15/VoCos25

    Correct?

    And then im lost from there, because you can't solve the next equation (y = y0 + v0yt - (1/2)gt2) for t. If I could solve for t=, then I could just put the 2 equations together and solve for Vo, Correct?

    Sorry, I'm a little confused and not quite grasping it.

    Thanks so much.
     
  7. Feb 17, 2013 #6
    bump?

    Ive tried subbing in the first equation (t=15/VoCos25) into the 2nd equation, but ended up with Vo^3 and it was just a mess. Im quite lost on this question still.

    Thanks in advance.
     
    Last edited: Feb 17, 2013
  8. Feb 17, 2013 #7

    cepheid

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    Why not? It's a quadratic. Have you heard of the quadratic formula?
     
  9. Feb 18, 2013 #8
    Never mind, I forgot to cancel out a V, ended up with 14.9m/s for my Vo, and I believe it works :).

    Thanks for the help :D
     
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