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Basketball question HELPP

  1. Oct 11, 2006 #1
    A basketball player shoots a ball at an angle of 60 with an initial velocity of 7.0 m/s. The ball was 2.0m above the ground when released and the basket was 3.00m high. How long was the ball in the air?

    Here is what i did so far:
    y-axis x-axis
    v1=7.0m/s x sin60 =6.06 m/s v1=7.0m/s x sin30 = 3.5m/s

    I used v2squared = v1squared +2ad
    since v2=0 0=6.06m/s squared + 2(-9.81m/s2)d

    To find time i used d=vt+1/2atsquared
    -1.87m=(6.06m/s)2t + 1/2(-9.81m/s2)t2
    after doing the quadratic formula i got t=1.49113
    Therefore the ball was in the air for 1.49113s/2 = 0.75s
    Should i divide it by 2 or noooo??? im not sure if its the maximum height...
  2. jcsd
  3. Oct 11, 2006 #2


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    Why would the y-velocity be 0 when the ball reaches the net?
  4. Oct 11, 2006 #3

    so v2 isn't zero in this case. i was supposing that v2 will be 0, when the ball hits the ground. So how would you proceed with this problem further?
  5. Oct 11, 2006 #4


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    I would write the equation of kinematics for y(t) and solve for the time when y=3. I will have to solve a quadratic equation in t. The lesser solution for t is the first time the ball got up to 3m (it hapened then the ball as ascending). The greater solution is what I'm looking for; it is the time when the ball is descending and on its way to a glorious 3-pointer.
  6. Oct 11, 2006 #5
    I'd think that your second equation is sufficient, except you'll want to "fix" it to either:

    [tex]d=d_0 + v_0t+\frac{1}{2}at^2[/tex]

    or else, if you're not familiar with having a starting position on the right side of that equation, simply use the vertical displacement the ball is going to have: 1.00 meters.

    Of course, you'll use the vertical component of velocity that you already found. Incidentally, I believe there will be two solutions for t; one corresponding to shooting up through the hoop, and the other for it falling down through the hoop.
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