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Projectile Motion Problem with stunt driver

  1. Dec 14, 2009 #1
    Stunt driver hired to do a scene where a motorcycle speeds horizontally off a 30m high cliff. There's a soft landing pad located 90m from the base. Ignore air resistance. (A) How fast must he go land on the soft padding? (B) the motorcycle has an acceleration of 5 m/s^2 on a dry road surface, he needs a certain distance from the edge to build up required speed. What is the distance?


    My attempt: (A) Y= Yo + Vxo(T) + 1/2 (g) (T^2) ---> 0= 30 + 0 + (-5) T^2 --->T=sq.rt 6 ---> T=2.44s (we use 10 instead of 9.8)

    Xf= Xo + Vxo(T) + 1/2 (Ax) T^2 ---> 90=0 + Vxo (2.44) +0 ---> Vxo = 36.8 m/s

    (B) X=Xo+VxoT + 1/2 (Ax)T ---> X-Xo= (36.8)(2.44) +0 ---> Displacement = 89.8 m



    Did I do this correct?
     
  2. jcsd
  3. Dec 14, 2009 #2
    Question is asking "build up required speed" ... So what distance he needs to come with horizontal velocity component of 36.8 m/s (which looks good) if he has acc of 5 m/s^2 ... That is before he jumps so the time he is in the air has nothing to do here.
     
  4. Dec 14, 2009 #3
    So would I use:


    (Vf)^2 = (Vxo)^2 + 2(Ax) (Displacement)

    0-36.8^2 = 10 (Displacement)
    1354.24/10 = Displacement = 135.42 m
     
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