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Basketball (Energy and its Conservation?)

  1. Nov 19, 2015 #1
    1. The problem statement, all variables and given/known data

    A basketball player shoots a 3-point shot. The 0.625 kg ball leaves the player's hands with a speed of 9m/s. It rises 2.4m before descending towards the basket. at the top of the ball's arc, what is its speed
    2. Relevant equations

    Im at a chapter named Energy and its conservation so im guessing i have to use these
    PE=mgh
    E= KE + PE
    KE(before)+PE(before)=KE(after)+PE(after)
    3. The attempt at a solution
    I also tried some other equations such as
    Velocity(final)^2= V(initial)^2 + 2a(d)
    V(f)= 9^2 + 2(9.8(don't know if it would be gravity))(2.4)
    (Please try to fix any wrong ideas that i have in my head)
     
  2. jcsd
  3. Nov 19, 2015 #2
    Do you know projectile motion? Essentially, do you realize that the velocity might have two components, in both x and y directions?
     
  4. Nov 19, 2015 #3
    Like vector addition?
     
  5. Nov 19, 2015 #4
  6. Nov 19, 2015 #5
    And that you might have to deal with the components separately, because the equations that you've used are only valid for 1 Dimensional motion (unless you write them in vector form, and you haven't done that).
     
  7. Nov 19, 2015 #6

    haruspex

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    That is true if one is to apply the SUVAT equations here. However, this is unnecessary and does not seem to fit with the intent of the question. It is most easily answered entirely in terms of work conservation, without considering separate components or vectors.
    @Chilling, define 'before' and 'after' in the equation you quote in the context of this question. What do you know about the KE and PE at those times?
     
  8. Nov 19, 2015 #7
    Actually my teachers answer was incorrect, I also checked with my classmates thanks for the help. The attempt I made was correct using Velocity(final)^2= V(initial)^2 + 2a(d).
     
  9. Nov 19, 2015 #8

    haruspex

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    Yes, that equation is right, and I'm glad to hear you got the right answer, but what you posted originally is wrong:
    That would give a speed greater than the initial speed.
     
  10. Nov 19, 2015 #9
    Oh I see i forgot the squared on velocity final
     
  11. Nov 19, 2015 #10

    haruspex

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    Funny - I didn't even notice that. There's another error.
     
  12. Nov 20, 2015 #11
    I'm sorry, I didn't pay attention to it being from energy conservation. I was thinking in terms of this, especially since it mentioned ball's ARC--
    Hmax = (u2(sinθ)2)/2g. (Formula for maximum height in a projectile)

    Hence find usinθ, i.e., y component of the velocity.

    Using (usinθ)2 + (ucosθ)2 = u2 we get ucosθ, i.e., the x component of the initial velocity, which is the only velocity that remains at the top of the ball's arc (since the y component of the velocity has become 0).
     
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