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Two-dimensional Kinematics

  1. Apr 25, 2014 #1
    1. The problem statement, all variables and given/known data

    A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp (Fig. 3–41). (a) With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5 m above the cars and the horizontal distance he must clear is 22 m. (b) If the ramp is now tilted upward, so that “takeoff angle” is 7.0° above the horizontal, what is the new minimum speed?


    2. Relevant equations
    [tex]y=y_0+v_0t+.5at^2[/tex]
    [tex]v^2=v_0^2+2a(y-y_0)[/tex]
    [tex]x=v_0t[/tex]


    3. The attempt at a solution

    The answer to part a is obviously 40m/s, but the answer to b eludes me.

    I need v, so I defined v_x in terms of v via the kinematic equation [tex]x=v_0t[/tex]

    and then solved it for t

    [tex]22=tcos\theta[/tex]

    [tex]t=\frac{22}{vcos\theta}[/tex]

    However, when I attempt to do this with v_y, I end up with a nasty quadratic

    [tex]-4.9t^2+tvsin\theta+1.5=0[/tex]

    and solving this for t does not seem to be advisable.
     
  2. jcsd
  3. Apr 25, 2014 #2

    SammyS

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    Hello, ciubba. Welcome to PF !

    Why do you consider that quadratic equation to be nasty ?

    You know that θ = 7° . Right ?

    You should probably solve the other equation for t, in terms of v, then substitute that back into the quadratic, so there's only one variable.
     
  4. Apr 25, 2014 #3
    I can't believe that never occurred to me-- I was going to solve them both for t, then set them equal to each other.

    The answer is 23.88, which rounds to 24m/s with significant digits. Thanks!
     
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