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A skateboarder shoots off a ramp with a velocity of 6.7 m/s, directed at an angle of 53° above the horizontal. The end of the ramp is 1.0 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp.

When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

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4 answers

  1. I keep seeing the 4.9 where did that come from?

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  2. I believe 4.9 is half of A=9.8 inside of the equation Y=Voy(t)+Ay(t2)/2

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  3. Why did the + in the equation switch to a -?

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  4. a)
    Vi= 6.7sin(53)
    Vi = 5.35

    V = Vi - gt
    at top 0 = Vi - g t
    so t at top = Vi/g
    t = .55
    y at top = 1 + Vi t - 4.9 t^2
    = 1 + 5.35(0.55) - 4.9(.55)^2
    y at top = 2.46

    b)
    u = 6.7 cos 53
    x = u t
    x = (6.7 cos 53)(.55)
    x = 2.22

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