Physics

A basketball player, standing near the basket to grab a rebound, jumps 73.6 cm vertically. How much time does the player spend in the bottom 15.4 cm of the jump?


Ok...after this I swear I'm done!
I just need someone to check my final answer, which is 0.0858 s.

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  1. He jumps .736m high, time in air..
    time in air falling > 1/2 9.8 t^2=.736
    time to fall .763-.154> 1/2 9.8 t2^2=.609

    t= sqrt ..736/4.9= .388sec
    t2= sqrt.119 = .345s
    t-t2= .0434
    Now double it for the time going up.
    time=.0868 seconds

    check my math, we are off on one digit.

  2. See, I use your method without rounding and now come up with 0.0852 s. So...now I'm a little stuck on which answer is the correct one.

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    posted by Lindsay

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