A long jumper approaches his takeoff board A with a horizontal velocity of 30ft/s. Find the vertical component (vy) of the velocity of his center of gravity at takeoff for him to make the jump if he lands at a spot 22ft from point A. What is the vertical rise of his center of gravity?

So when I started working on it I made separate equations for the x and y components but it didn't turn out right. Is it even necessary to make separate equations?

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  1. Get his time in the air, T, from the horizontal distance and velocity component.

    T = 22/30 = 0.7333 s

    The vertical velocity component must decelerate to zero in half that time (when maximum elevation is reached)

    Vy = g*(T/2)= 3.59 ft/s

    The vertical rise of the CG is given by
    (g/2)*(T/2)^2 = H
    or (Vy/2)*(T/2)
    They give the same result.

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  2. I see how this would be correct but the book says the answer is 11.81ft/s and the vertical rise is 2.16 ft

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  3. His methods for finding the answer is correct. The only problem is that he uses 9.81 for acceleration by gravity and in this particular problem all dimensions are in ft. So use 32.2 ft/sec and you will find the right answer.

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