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Physics

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A basketball player, standing near the basket to grab a rebound, jumps 65.6 cm vertically. How much time does the player spend in the top 10.7 cm of his jump

  • Physics -

    We will treat the player as a point mass, and refer to where his center of mass is located. To raise the center of mass by H = 0.656 m, the TOTAL time spent in the air must be 2*sqrt(2H/g) = 0.731 s . Half of that time is spent going up and half coming down.

    The time to fall 0.107 m from the max height of 0.656 m to 0.549 m is
    t = sqrt(2*0.107)/g) = 0.1477 s
    The total time spent above that altitude is twice that, or 0.295 s.

  • Physics -

    top at .656 meters
    get Vi, initial speed from conservation of energy
    (1/2) m Vi^2 = m (9.81)(.656)
    Vi^2 = 2 (9.81)(.656)
    Vi = 3.59 m/s
    Now do a new problem, how fast is he going at h = .549 meters?
    h = .549
    (1/2) m v^2 + mgh = (1/2) m Vi^2
    v^2 = Vi^2 - 2 g h
    v^2 = (3.59)^2 - 2*9.81*.549
    v = 1.45 meters/second
    That is our initial condition for the top part
    How long to top from there?(0 velocity up)
    0 = 1.45 - 9.81 t
    t = .148 s
    total up and down 2 times t = .297 s

  • Physics -

    How long does he stay in the air?
    Well, his jump-up time is the same as if he had fallen from .656m

    s = 1/2 at^2
    .656 = 4.9t^2
    t = .3659 sec
    so, he rose for .3659s and then fell for .3659s

    His velocity at landing is the same as it was at takeoff:
    v = at
    = 9.9 * .3659
    = 3.5858 m/s

    So, the equation of motion giving his height is

    h = 3.5858t - 4.9 t^2
    Note that h=0 at t=0 and t=.7318

    Now, how long was he above 65.6-10.7 = 54.9cm?

    Translate the graph down .549 and solve
    3.5857t - 4.9 t^2 - .549 = 0

    He was above 54.9cm from t=.2181 to t=.5136, or a total of .2955 seconds.

    ute problem.

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