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Projectile basketball

  1. Oct 20, 2007 #1

    ~christina~

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    [SOLVED] projectile basketball

    1. The problem statement, all variables and given/known data
    A 40kg basketball player executes a impressive shot. The play-by-play commentator describing the live action says that the student launched herself at an angle of 35 degrees releasing the ball when she was a horizontal distance of 4.31m from the center of the basket.

    a) if the ball was released 1.83m above the floor and the basket is a height of 3.05m above the floor, what was the velocity of the ball when it was released?

    b)what was the maximum height of the ball above the floor?

    c) what is the velocity of the ball as it enters the basket?

    d) What was teh time of flight of the ball?

    e) how does the magnitude of the force that the player uses to accelerate the basketball compare (that is less than, equal to, or greater than) with the magnitude of the force that the ball exert on the player?
    Use correct physics principles to explain your reasoning..



    2. Relevant equations
    Vxf= vxi + axt

    Xf= xi + vxi*t + 1/2 axt^2


    3. The attempt at a solution


    Well I know that the
    angle = 35 deg
    sx= 4.31m
    sy= 3.05m- 1.83m= 1.22m

    Vo= ?
    t= ?

    I need Vo so I plugged in...

    Sx= Sox +vx*t + 0.5axt^2 (Sox= 0 and no acceleration in x direction a= 0)
    Sx= vx*t
    Vx= vo cos theta

    Sx= vo (cos theta)*t

    Sy= Soy+ Voy *t + 0.5gt^2 (Soy= 0 g= -9.8)
    Voy= Vo sin theta
    Sy= Vo (sin theta) *t - 4.9t^2


    since I need the initial velocity...
    I think I plug into the other equation the equation for the distance in the y direcction ...

    Sx= vo *cos theta *t
    t= Sx/ (vo*cos theta)

    Sy= vo sin theta*t - 4.9 t^2

    Sy= vo sin theta (Sx/ vo cos theta)- 4.9 (Sx/ vo cos theta)^2

    Sy= (tan theta *Sx) - 4.9 (Sx/vo cos theta)^2

    Sy- tan theta* Sox = -4.9 ( Sx/ vo cos theta) ^2

    [tex]\sqrt{} (Sy- tan theta*Sox) / -4.9[/tex]= Sx/ vo cos theta

    Vo cos theta = Sx/ ([tex]\sqrt{} (Sy- tan theta*Sox) / -4.9[/tex])

    Vo = need
    theta= 35 deg (assuming same angle when player jumps)
    Sx= 4.31m
    Sy= 1.22m (after subtracting the different heights)

    Vo cos 35 = 4.31m / ([tex]\sqrt{} (1.22m- tan 35*4.31m) / -4.9[/tex])

    Vo cos 35= 7.12

    Vo= 8.69m/s^2

    Well I think I did my math correctly...could someone check it for me?

    b.)

    Max height of the ball....

    Vy= 0 at max height...

    time at max height is = ?

    Well I think ..

    Vy= Vo sin theta - gt
    0= Vo sin theta - gt

    t= Vo sin theta/ g (since negatives cancel out)

    t= (8.69m/s^2) (sin 35)/ 9.8

    Then t= 0.5086s to reach max height

    I don't know if this is fine but...
    plugging that into y distance equation..

    Sy= Soy+ Vo sin theta*t - 4.9 t*2

    Soy= 0
    Vo= 8.69 m/s^2
    t= .5086s

    Sy= 4.98(0.5086s)- 4.9 (0.5086s)

    Sy= 1.265m (max height if only including the part that I calculated after subtracting the difference in the original height)

    Sy actual= 1.265 + ?= ( I really don't know what I add [can't figure it out])
    c.) Assuming the previous parts were okay..( I need help with the above to find out what to add)
    velocity of the ball as it enters basket....

    I'm not sure how to find his...


    Can someone check and help me out...?
    Thanks Very much
     
  2. jcsd
  3. Oct 21, 2007 #2

    cristo

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    Your algebra here is correct, so presuming your arithmetic is correct, this this is the right answer.

    Ok, good.

    Well, the "initial" velocity you are using is the velocity when the player has a vertical height of 1.83 metres. So, in fact, your Sy0 should not be zero, but should be 1.83m.
    Well, what more information do you know about this scenario? What is the vertical height? The horizontal height?

     
  4. Oct 22, 2007 #3

    ~christina~

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    For some reason I couldn't edit the original post so I'm uploading the image I drew for the problem...

    [​IMG]

    The reason I didn't use 1.83m was because I subtracted the 2 heights of 3.05m and 1.83m so that I ended up with the begining height as 0 and the end height as 1.22m so it would be easier to deal with. And I couldn't decide which number I had to add to the max height I found to get the original real max height in relation to the ground.

    I really need help on how to find the velocity of the ball as it enters the basket...

    I have a test on wed..and i neeed to know how to find this because they'll definately be a problem like this on there..

    PLEASE HELP:cry:
     
  5. Oct 22, 2007 #4

    Doc Al

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    You need to add back the 1.83 m she started with.

    (Parts a and b look good.)
    You'll use the same equation for velocity as used in part b. Figure out the time from what you did for part a. Hint: Only the vertical component of velocity changes.
     
  6. Oct 22, 2007 #5

    ~christina~

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    For b) ymax 1.265m + 1.83m= 3.095m

    c) v as basketball enters basket
    t enter = ?
    Vy= ?
    Vx= vo cos theta= 8.69cos 35= 7.12m/s^2 (stays constant in x direction)

    t to get to basket= ?
    Sx= Vo cos theta *t
    4.31m= 8.69 cos 35*t

    t= 0.605s

    Vfy= Vosin theta - gt

    Vfy= 8.69 sin 35( 0.605s) - 9.8 (0.605s)

    Vfy= 2.02m/s

    at the basket the velocity is
    V= [tex]\sqrt{}(7.12)^2 + (2.02)^2[/tex]

    V= 7.40m/s


    d.) time of flight of ball
    ~do they mean the time it takes for the ball to reach 0?

    I think that it is from the distance to the max height in the x direction and just times that by 2 but I'm not sure...

    I think that I go and take the time it takes for the ball to reach the max height and plug that into the x distance equation then multiply that distance by 2...

    Well....doing that .. (assuming it's alright)

    from before
    tmax= 0.5086s

    then ..
    Sx= Sox + vxt => there is no acceleration in x direction and thus a=0 canceling out 0.5axt^2

    Sx= 0 + 8.69cos 35 (0.5086s)

    Sx= 3.62m ===> not sure if this is correct but if it is..

    Sx half = 3.62m
    Sx total distance = 3.62m *2= 7.24m

    Sxtotal= 7.24m

    time of flight ...using the x distance equation

    Sx= Sox + vxt

    7.24m= 0 + 8.69m/s cos 35 (t)

    t= 1.0171s

    I'm not even sure why I went through all that for the time of flight since I already got the time at max height and could have just multiplied it by 2 and get the same answer..however I didn't see that...oh well...

    e) is the force exerted by the basketball player on the ball greater or less than the force exerted by the basketball on the player?

    I think it is the same since Newton's 3rd law is that every action has a equal and opposite reaction and thus the basketball player exerts a force on the ball and the ball exerts a equal force back.

    what I'm not sure about is why the basketball moves...is it because the basketball player exerts a greater force on the ball than the ball exerts on the person?
    (this has me confused since I would logically think that the reason a thing doesn't move is b/c one places a equal and opposite force on a object but if it moves one way then shouldn't it be not equal at least at the begining when the object moves?)

    Thanks :smile:
     
    Last edited: Oct 22, 2007
  7. Oct 22, 2007 #6

    Doc Al

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    Double check your calculation for Vfy.

    Time of flight means the total time the ball is in the air from the moment it leaves her hand to the moment it goes in the basket. (You already figured this time out in doing part c!)

    Only if the initial and final positions are at the same height will the time to reach the maximum height be exactly half of the total time. (Your diagram is somewhat misleading.)

    Absolutely.

    The ball accelerates because there's a net force on it. Realize that the force pairs in Newton's 3rd law act on different bodies. In determining the motion of the ball, we care about the forces exerted on the ball. It's true, by Newton III, that the ball also exerts an equal and opposite force on the player, but that's got nothing to do with the motion of the ball.
     
  8. Oct 22, 2007 #7

    ~christina~

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    For b) ymax 1.265m + 1.83m= 3.095m

    c) v as basketball enters basket
    t enter = ?
    Vy= ?
    Vx= vo cos theta= 8.69cos 35= 7.12m/s^2 (stays constant in x direction)

    t to get to basket= ?
    Sx= Vo cos theta *t
    4.31m= 8.69 cos 35*t

    t= 0.605s

    Vfy= Vosin theta - gt

    Vfy= 8.69 sin 35( 0.605s) - 9.8 (0.605s)

    Vfy= -2.91m/s

    at the basket the velocity is
    V= V= [tex]\sqrt{}(7.12)^2 + (-2.91)^2[/tex]

    V= 7.69 m/s

    Thanks for catching that Doc Al, I think it should be correct now

    I'm kind of confused..is it that acceleration is always -9.8 if it's g or is it 9.8 b/c sometimes I don't use the negative and that would give me a different answer.

    Thanks again Doc Al:smile:
     
  9. Oct 22, 2007 #8

    cristo

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    The sign of the acceleration, like the sign of any other quantity, depends upon your choice of coordinate system. In this question, you have defined upwards as positive, and thus everything up has a positive sign, and everything down (like the acceleration) has a negative sign. There are some questions where it is more useful to define the coordinate system the other way around. In that case, downwards would be positive, and so the acceleration would have a positive sign.
     
  10. Oct 22, 2007 #9

    ~christina~

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    Thank you cristo for clearing that up for me :smile:
     
  11. Oct 22, 2007 #10

    Doc Al

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    Looks good.

    For free projectiles, the acceleration is always 9.8 m/s^2 downward. If you are using a standard sign convention (up = +, down = -) then the acceleration will be -9.8 m/s^2.

    Sometimes it's easier to take down as positive as a shortcut. For example: How far does a ball fall in 2 seconds? I'd just use d = 1/2gt^2 = 4.9*4= 19.6m. I know it falls down, so I don't care about the sign. Done formally (using the standard convention) I'd have to say: y = y0 -1/2gt^2 = -19.6 m. Same difference.

    In a complicated problem, stick with the sign convention. :wink:

    FYI: In common usage, g stands for the magnitude of the acceleration, thus g = 9.8 m/s^2. It's up to you to put in the sign as needed.

    (Looks like cristo beat me too it! :smile:)
     
  12. Oct 22, 2007 #11

    ~christina~

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    Okies...I get it

    cristo did beat you Doc Al but tis strange..I swear I didn't see him online and suddenly a reply appears...:bugeye:

    Thanks again for your help :smile:
     
  13. Oct 22, 2007 #12

    ~christina~

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    For b) ymax 1.265m + 1.83m= 3.095m

    c) v as basketball enters basket
    t enter = ?
    Vy= ?
    Vx= vo cos theta= 8.69cos 35= 7.12m/s^2 (stays constant in x direction)

    t to get to basket= ?
    Sx= Vo cos theta *t
    4.31m= 8.69 cos 35*t

    t= 0.605s

    Vfy= Vosin theta - gt

    Vfy= 8.69 sin 35( 0.605s) - 9.8 (0.605s)



    Vfy= -2.91m/s

    at the basket the velocity is
    V= V= [tex]\sqrt{}(7.12)^2 + (-2.91)^2[/tex]

    V= 7.69 m/s


    I made a error on that (multiplied the time in for some strange reason) and I found that...but if I use the g as -9.8 for the g or use positive 9.8 as the g I get 2 different answers...(point of confusion before)
    Vfy= Vosin theta - gt

    Vfy= 8.69 sin 35 - 9.8 (0.605s)= -.93m/s

    with the possitive...

    Vfy= 8.69 sin 35 + 9.8 (0.605s)= 10.90m/s


    Now I'm really confused...
     
  14. Oct 22, 2007 #13

    cristo

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    Hmm, can't believe neither of us spotted that mistake!

    Well, you will get two different answers, as they are two different equations. Your first equation, however, is the correct one. It's general form is v=u+at, so v=u-gt. So, the answer of -0.93m/s for the final vertical velocity is correct.
     
  15. Oct 22, 2007 #14

    ~christina~

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    Thanks Cristo :smile:
     
  16. Oct 22, 2007 #15

    cristo

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    You're welcome!
     
  17. Oct 22, 2007 #16

    Doc Al

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    Me neither! :redface:
     
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