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Homework Help: Ramp angle and landing velocity

  1. Jun 10, 2010 #1
    1. A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcycle. There is a 15. m drop and the horizontal distance originally planned was 60. m but it turns out the canyon is really 66.3 m across. If he desires a 3.0-second flight time, what is the correct angle for his landing ramp (deg, positive angle below horizontal) AND landing velocity?

    2. magnitude= sqrt(Vf^2 + Vi^2), etc

    3. I've calculated the launch speed to be 24.1 m/s and the launch angle to be 23.7 deg. I've tried using the formula 2. to solve for the landing velocity by figuring Vf 15sin(23.7)-(9.81)(3)= -23.4. plugging into the formula, sqrt(-23^2 + 24.1^2)= 33.6 and this is incorrect. I've exhausted myself trying to figure these last 2 parts, and I'm now stumped! Thanks in advance. I also added a visual that may help.

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    Last edited: Jun 10, 2010
  2. jcsd
  3. Jun 10, 2010 #2


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    this is correct.

    Recheck this as I do not -23.4. Also remember this will give you the vertical component, which must be added (vector wise) to the horizontal component of velocity which is not 24.1 m/s.
    Last edited: Jun 10, 2010
  4. Jun 10, 2010 #3
    I just rechecked the math and still came out with -23.4. Would the horizontal component be 22.1 from dividing 66.3/3 ? Also, how can the landing angle be calculated? Sorry...confused
  5. Jun 10, 2010 #4


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    24.1sin(23.7) - 9.81(3)

    yes the horizontal component would be 22.1 m/s.
  6. Jun 10, 2010 #5
    Right, wrong number. I figured out the ramp angle by dividing -19.7 by 22.1 and took the reverse tan to get 41.7 degrees. I the got the velocity by using sqrt(19.7^2 + 22.1^2) to get 29.6 m/s.

    Thanks for your help rock.freak667, I do appreciate it!
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