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Motorcycle stunt-rider mechanics problem

  1. May 17, 2010 #1
    1. The problem statement, all variables and given/known data
    A motorcycle stunt-rider moving horizontally takes off from a point 1.25 m above the ground,
    landing 10 m away as shown.


    2. Relevant equations

    3. The attempt at a solution
    using tan [tex]\alpha[/tex] = [tex]\frac{Opp}{Adj}[/tex]
    [tex]\alpha[/tex] = 7.1
    then using v2=u2+2as
    let a = g = 10
    s = 1.25(since we are calculating vertical)
    u2 = 25
    u = 5
    but u is Xsin [tex]\alpha[/tex]
    therefore x = [tex]\frac{5}{sin\alpha}[/tex]
    x = 40 ms-1
    no the answer is 20ms-1
    what is wrong?
  2. jcsd
  3. May 17, 2010 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Re: mechanics

    Your equation is not correct. The correct equation is:

    [tex]h = v_0\sin{\alpha}t - \frac{1}{2}gt^2[/tex]

    where h = change in vertical position = -1.25 m.

    The angle of launch is 0 degrees. Work out the time of flight. From that you can work out its horizontal speed (it travels 10 m in that time).

  4. May 18, 2010 #3
    Re: mechanics

    i have found it
    thx bro for stating the angle of launch
    i always thinking that for no angle is given
    we have to find the angle by ourselves using trigo ..
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