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Physics

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A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.

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..(I got 40 m/s)

B)If the ramp is now tilted upward, so that "takeoff angle" is 7.0∘ above the horizontal, what is the new minimum speed? (I cant figure out)

  • Physics - ,

    a= 37m/s
    b= 20m/s

  • Physics - ,

    Okay, I was having problems with this and tried twisted the numbers around, not sure if right but it could be.

    A) Vi=39.8 m/s

    B)Y=Yi+VyiT+(1/2)a(t^2)
    use X=Xo+VxiT
    to make T=x/Vxi
    This way you use substitution
    Y=Yi+Vyi(X/Vxi)+(1/2)a(X/Vxi)^2

    When you look at the vectors you can relate the angle and the two velocity vectors using tan(7)=(Vyi/Vxi)
    Substitute that back in and you get
    Y=Yi+tan(7)X+(1/2)a(X/Vxi)^2
    Y=final Y so Y=-1.5m
    X=final X so 22m
    a=-9.8m/s^2

    Using all of this you can find Vxi and from there get the other component as well as the answer

    I find that the answer is...
    Vxi=23.76m/s
    Vyi-2.92m/s
    Vinitial=23.94m/s

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