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Homework Help: Projectile basketball player

  1. Oct 22, 2007 #1


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    [SOLVED] projectile basketball player...

    1. The problem statement, all variables and given/known data
    In a jump to dunk a ball, a basketball star has a flight time of 0.852s (hang time)
    His motion through space can be modeled as that of a particle at a point called his center of mass. His center of mass is at elevation of 1.02m when he leaves the floor and is at elevation 0.900m when he touches down again.

    a) if he leaves the floor at an angle of 50.8 deg to the horizontal, what is the horizontal distance that he covers in the jump

    b) determine the magnitude of the velocity at the instance he leaves the floor

    c) what is his velocity, magnitude and direction at the instance he returns to the floor?

    I drew a picture
    http://img89.imageshack.us/img89/876/59145339rq6.th.jpg [Broken]
    2. Relevant equations
    kinematic equations

    Vf= Vi + at

    Vx= Vo cos theta

    Vy= Vo sin theta

    Sx= Sox+ Vxt + 0.5 a t^2

    3. The attempt at a solution

    Well I was confused first of all about how I solve this since the initial height is higher than the final height so I guess I wouldnt' be able to just draw a line over to the other side of the trajectory to say that the initial height= 0.

    Would I have to use the heights as is given?
    such as the Soy= 1.85m and Sy= 0.900m?

    (I just want to get this clear ..but I'll post my work after this assuming this is correct unless someone else says otherwise)

    a.) to find the distance covered in the jump I think I would use the distance equation for the x component...

    Sx= Sox + Vxt + 0.5axt

    since a= 0 in x direction...
    Vx= Vo cos theta

    however for that you'd need Vo so I guess I need to plug into the y component of distanc to find the Vo...

    Sy= Soy + Vyt+ 0.5at^2

    Sy= 0.900m
    Soy= 1.02m

    Vy= Vo sin theta

    a= -9.8
    theta= 50.8 deg
    t= 0.852s

    Sy= Soy + Vyt+ 0.5at^2

    0.900m= 1.85m + Vo sin (50.8) (0.852s) - 4.9 (0.852)^2

    -.12= .660 Vo - 3.56

    3.44 = .660 Vo

    Vo= 5.21m/s

    Since have Vo...plug into the x component of the distance equation

    Sx= Sox + Vxt + 0.5 at^2
    a= 0
    Sox= 0
    Sx= ?
    Vx= Vo cos theta => 5.21m/s cos 50.8 = 3.29 m/s
    Vo= 5.21 m/s
    theta= 50.8 deg
    t= 0.852s

    Sx= 3.29m/s (0.852s)

    Sx= 2.80 m ===> is it just me or this seems sort of small in x distance..:confused:

    basically need a check on how things are going here
    Thanks :smile:
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 23, 2007 #2


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    everything looks good to me.
  4. Oct 23, 2007 #3


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    b) determine the magnitude of his velocity at the instant he leaves the floor.

    I think I found this in part a (Vo) since I had to plug that into the equation for the x distance so would it be ..


    Vx= 5.21m/s cos 50.8= 3.29m/s

    Vy= 5.21m/s sin 50.8= 4.04m/s


    Vtotal= [tex]\sqrt{} (3.29m/s)^2 + (4.04m/s)^2[/tex]

    Vtot= 5.21m/s ==> wait..this is the same as Vo..haha.. slaps forehead* => or is it??


    c) what is his velocity, magnitude and direction , at the instance he returns to the floor?

    t= 0.852 s
    Vo= 5.21m/s
    Sx= 2.80m
    Sy= 0.900m
    Soy= 1.02m
    theta= 50.8

    x component stays same...
    Vx= Vocos theta
    Vx= 5.21cos 50.8= 3.29m/s

    Vx= 3.29m/s

    Vfy= Voy + ayt

    Vfy= Vosin theta - 9.8(0.852s)
    Vfy= 5.21sin 50.8 - 8.35

    Vfy= -4.31m/s or just 4.31m/s ==========> is this fine? I guess even though I got a

    negative I can change it to a possitive though but I'm not sure.

    Vtot= [tex]\sqrt{} (3.29m/s)^2 + (-4.31)^2[/tex]

    Vtot= 5.21

  5. Oct 23, 2007 #4

    Doc Al

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    Staff: Mentor

    Of course it's the same. :smile:

    Why change it? He's coming down, so Vfy should be negative. (Remember that you need to find direction as well as speed.)

    Check your arithmetic on that last step.
  6. Oct 23, 2007 #5


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    So vfy= -4.31m/s

    and in the end I get after checking the math...It isn't the same..it's

    V= 5.42m/s !

    (more than the original and that makes sense sinc he's lower than the initial height thus the velocity is greater => I think that's how I'm supposed to think about that)

    however for the direction...I was thinking about that..would it mean I have to find the angle??

    I would think that entails the finding of the angle with tan I'm not quite sure about that...
    not sure how the velocity component's look and which is which to plug into the

    tan theta = y/x

    would it be the velocity total that I found over the y component of the velocity?

    tan theta= 5.42/ -4.31
    tan theta= -1.25

    theta= -51.5 ?

    How is this?

    Thank You Doc Al :smile:
  7. Oct 23, 2007 #6

    Doc Al

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    No, it would be: tan theta = Vy/Vx
  8. Oct 23, 2007 #7


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    I get is now...

    so it would be

    tan= Vy/Vx

    tan= -4.31/ 3.29= -1.31

    theta= -52.6 deg
    Last edited: Oct 23, 2007
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