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Projectile motion motorcycle jump over cliff

  1. Sep 29, 2009 #1
    a motorcyclist jumps off a cliff inclined at 53.0 degrees over a river that is 40.0m wide. the far bank is 15.0m lower than the edge of the take off ramp. the river itself is 100m below the ramp. Ignore air resistance. What should his speed be at the top of the ramp to just make it to the edge of the far bank?

    given:
    theta = 53.0 deg.
    d = 40.0 m
    change(y) = 15.0 m

    I'm just not sure what equation of projectile motion to use. Time is not really a factor in this... and I am confused because the variable t is in all the equations and where I don't know initial velocity, it leaves me with two unknown variables. Can someone just point me in the right direction, help me figure out which equations to use. Thank you so much.
     
  2. jcsd
  3. Sep 29, 2009 #2

    berkeman

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    Staff: Mentor

    You end up with two simultaneous equations, so you can solve for both unknowns. One equation is for the horizontal motion, which has a constant velocity. The other equation is for the vertical motion as a function of time, and has a parabolic term in it due to the constant acceleration of gravity changing the vertical velocity as a function of time. Does that help?
     
  4. Sep 29, 2009 #3
    so i use x(t) = v(x)t + x(0) and the other equation is y = y(0) + v(0y)t - 1/2gt^2 ? even when i substitute known variables into these i have no idea what to do with them
     
  5. Sep 29, 2009 #4

    berkeman

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    Staff: Mentor

    You should clean up those starting equations a bit first. The velocity in the first equation should be v(t), for example, not v(x) (Quiz Question -- how come?). And you should be more explicit about the components to avoid what you are writing in the second equation "V(0Y)t" is confusing and potentially wrong. Try using Vx(t) and Vy(t) for the velocity names.

    Now, you *do* have starting and ending coordinates, and and starting Vx(0) and Vy(0) values. Start writing those equations...
     
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