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Basketball question

  1. Oct 9, 2006 #1
    4.54) A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as in Figure P4.54. If he shoots the ball at a 40.0 degree angle with the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the the backboard? The basket height is 3.05 m.


    Again I drew a diagram. I could subtract 3.05 from 2.00 to give me 1.05m. So now I figure I can use equations

    vxi = (vi)(cos)(angle)
    and
    vyi = (vi)(sin)(angle)

    I have as a result:

    10 = (vxi)(t)
    and
    1.05 = (vyi)(t) - 4.9 t^2

    As you can see, I'm missing a few things. but I go through the motions of multiply through what i can.

    for x: 10 = (vi)(cos)(40)

    for y: 1.05 = (vi)(sin)(40)t - 4.9t^2

    =

    10 =(vi)(0.76604)

    =

    10/0.76604 = 13.05 (Initional velocity)

    I than use the intital velocity and plug it into the y component equation. 1.05 = (13.05)(sin)(40)t - 4.9t^2

    1.05 = (8.38834)t - 4.9t^2

    =

    -4.9t^2+8.38t-1.05

    now that we have a quadratic equation, I put it through a quadratic formula to give me 1.5 seconds.

    so the equation I use for intitial velocity is x = v(t).

    10 = vxi (1.5) = 6.66 m/s

    and

    1.05 = vyi (1.5) - 4.9 (1.5)^2 = 8.05 m/s

    so once I have those two velocities, I use a^2+b^2=c^2 to find the hypoteuse which yields 10.4478 m/s for my initial velocity.



    Just to let you know, I am very new to physics, in fact this is teh first college physics course I am taking, so for all I know, all my answers could be wrong. If you guys can let em know the right way about going through problems, and tips or tricks, it would be much appreciated.

    If you guys can double check my answers and let me know, I would really appreciate that as well :)

    take care and thanks very much!

    Neeraj
     
  2. jcsd
  3. Oct 9, 2006 #2

    OlderDan

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    What happened to the (t) I added to your quote? Looks to me like you lost it.
     
  4. Oct 10, 2006 #3
    ooops, right you are, it was a typo. But does the answer look right? Is the rest of my approach right?
     
  5. Oct 10, 2006 #4

    OlderDan

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    If you dropped the (t), then your 13.05 number is not right, and since you used it, what follows cannot be right. Go through it yourself with the correction and see where it leads. A lot of what you did seems to be on the right track, but that (t) will change things.
     
  6. Oct 10, 2006 #5
    oh I see!!!! I was close though!!! lol. I will make the necessary corrections and post them soon. Can't express how much I appreciate your help Dan!!!

    Thanks a million!

    Neeraj
     
  7. Oct 12, 2006 #6
    made a mistake with the x-component
     
    Last edited: Oct 12, 2006
  8. Oct 12, 2006 #7
    Ooops, I think I made a grave error after I founf my vxi = 13.05 m/s.

    I assumed I could take that velocity and plug it into my equation for the y-component, which I now realize is totally wrong. Both x and y components should be solved seperately. I mistook 13.05 to be vi instead of vxi. Know this, how can I find the initial velocity for my y-component?

    any ideas?
     
  9. Oct 12, 2006 #8
    k, I think I may have figured out my error. Let me know if this sounds good. :)



    4.54) A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as in Figure P4.54. If he shoots the ball at a 40.0 degree angle with the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the the backboard? The basket height is 3.05 m.


    Again I drew a diagram. I could subtract 3.05 from 2.00 to give me 1.05m. So now I figure I can use equations

    vxi = (vi)(cos)(angle)
    and
    vyi = (vi)(sin)(angle)

    I have as a result:

    10 = (vxi)(t)
    and
    1.05 = (vyi)(t) - 4.9 t^2

    As you can see, I'm missing a few things. but I go through the motions of multiply through what i can.

    for x: 10 = (vi)(cos)(40)

    for y: 1.05 = (vi)(sin)(40)t - 4.9t^2

    =

    10 =(vi)(0.76604)

    =

    10/0.76604 = 13.05 (Initional velocity)

    I than use the intital velocity and plug it into x=(vi)(cos)(angle)(t)

    10 = 13.05cos 40 \t

    10=(9.9996)t

    t = 1.000 sec

    =

    so now that I have my time, I plug that into my y-component equation:

    1.05 = vi sin 40 (1.000) - 4.9 (1.000)^2

    which gives me 9.2566 m/s

    Now that I have my vxi and vyi. I can plag those into pithegora's theorum to find my hypoteneuse.

    (13.05)^2 = (9.2566)^2 = 255.987 ( I take the square root and get....)

    15.9995 m/s for my initial velocity

    (does that sound right) ?
     
  10. Oct 12, 2006 #9

    OlderDan

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    Go back to the beginning and write your equations for the x and y positions versus time. You can solve the x equation for (t) and substitute the exprssion for (t) into the y equation. You will then have a relationship between the two initial velocity components. If you had a second relationship between these components, you could solve for them. The problem gives you the information you need to get this second relationship. What is it?
     
  11. Oct 12, 2006 #10
    would it be this?


    10/0.76604 = 13.05 (Initional velocity)

    I than use the intital velocity and plug it into x=(vi)(cos)(angle)(t)

    10 = 13.05cos 40 \t

    10=(9.9996)t

    t = 1.000 sec

    =

    so now that I have my time, I plug that into my y-component equation:

    1.05 = vi sin 40 (1.000) - 4.9 (1.000)^2

    which gives me 9.2566 m/s

    Now that I have my vxi and vyi. I can plag those into pithegora's theorum to find my hypoteneuse.

    (13.05)^2 = (9.2566)^2 = 255.987 ( I take the square root and get....)

    15.9995 m/s for my initial velocity

    (does that sound right) ?

    The second relationship you speak of is the Initial velocity when you solve for the hypotenuese, right?)

    *
     
  12. Oct 12, 2006 #11

    OlderDan

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    No. You can't find the initial x velocity from just the one equation. Do what I said in my previous post.
     
  13. Oct 12, 2006 #12
    ooop! I see my error. Just realized I can't have 13.05 as an inital x velocity.

    So going back to the begining, I have the two x and y equations:

    10 = vxi (cos) (40) t

    and

    1.05 = vyi (t) - 4.9t^2


    the way I see it, I can find time from the y equation, if I assume that vyi is 0 as that's the apex of the hight if the ball. so I'm left with...

    1.05= 0- 4.9t^2

    1.04/4.9 = t^2

    t = 0.4629 x 2

    t = 0.9258

    can I plug that into my x equation to find vxi? Or did I make another mistake? (I know you told me I can get time from the x equation, but it looks like I have 2 unknowns there. I have to hand it to you, you're being very patient with me, I really appreciate it Dan! Thanks!)
     
  14. Oct 12, 2006 #13
    oh wait. that's wrong, I can't have vyi = 0. That wouldn't make sense. sorry.
     
  15. Oct 12, 2006 #14

    OlderDan

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    Good for you. You are going to have to find those two relationships between the velocity components I mentioned earlier. There is no avoiding it.
     
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