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A man on a motorcycle plans to make a jump as shown in the figure. If he leaves the ramp with a speed of 29.5 m/s and has a speed of 28.0 m/s at the top of his trajectory, determine his maximum height (h) above the end of the ramp. Ignore friction and air resistance.

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Jesse
1 answer

  1. I do not see any figure.

    His speed at the top is his horizontal speed, u, for the whole trip

    u = 28 m/s

    at takeoff v^2 = Vi^2 + u^2
    where v is total speed
    and Vi is initial speed up
    so
    29.5^2 = 28^2 + Vi^2
    Vi = 9.29 m/s
    that is all I need to do the vertical problem
    easy with energy
    m g h = .5 m Vi^2
    h = .5 Vi^2/9.81
    h = 4.4 meters

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    Damon

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