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Projectile motion skateboarder throws a ball

  1. Jun 2, 2006 #1
    A student skateboarding at a constant speed of 5m/s throws ball vertically into the air at 4m/s. She catches the ball as it returns to the ground. Let the ball be released from her hand at time t=0, the moment she passes the marker.

    A) What is the initial velocity of the ball?

    B) How long is the ball in the air (time of flight)?

    C) How far past the marker is the student when she catches the ball (Range)? (Assume she catches the ball at the same height she throws it)

    I'm sure that the initial velocity is 5i + 4j.

    And i can get to,

    a = -9.8j
    v = -9.8t j + c
    v(0) = c = ucos90 i + usin90 j
    v = ucos90 i + (usin90 - 9.8t) j
    (angle is 90 as thrown straight up, i hope)

    In this case Im not sure how to incorporate the initial velocity into it
    Im sure if i got A) i could get the rest.

    Hopefully someone can help.
     
    Last edited: Jun 2, 2006
  2. jcsd
  3. Jun 2, 2006 #2
    The initial angle of projection is going to be [itex] arctan (4/5) [/itex] is it not? Since it has the initial velocity of 5i + 4j. If the angle was 90, then the person would not be moving sideways.

    To work out the time of flight, consider the displacement rather than the velocity and solve for time.

    If you have the time, then it is simply time*5 to get the horizontal distance from the marker.
     
  4. Jun 2, 2006 #3

    Hootenanny

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    The inital velocity is indeed 5i + 4j. I would leave it in the form as it makes answering the rest of the question easier. You have a velocity in unit vector form, i is horizontal and j is vertical. Consider these two components totally independantly.

    As you correctly said in the j direction, there is only one acceleration (that due to gravity). Knowing the intial vertical velocity, the acceleration and the final displacement of the ball can you use kinematic equations to find the flight time?

    ~H
     
  5. Jun 2, 2006 #4
    The equations we have been doing in class have us using the initial velocity in cartesian form, ie. 5m/s.

    Therefore I am having trouble using the initial velocity in the form it is in.

    maybe this...

    v(0) = c = ucos90 i + usin90 j + 5i + 4j
    v = (ucos90 +5)i + (usin90 - 9.8t + 4)j

    and maybe u would equal the mod of 5i + 4j, which would equal root 41 m/s.

    ???

    Once i find an equation for the position vector, r(t), i should be fine.
     
    Last edited: Jun 2, 2006
  6. Jun 2, 2006 #5

    Hootenanny

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    Okay, I'll set up an equation for you.

    In the vertical direction;

    vi = 4 m.s-1
    a = - 9.81 m.s-2
    s = 0 m

    Now using the kinematic equation;

    [tex]s = v_{i}t + \frac{1}{2}at^2[/tex]

    Can you find the time of flight (t)?

    ~H
     
  7. Jun 2, 2006 #6

    HallsofIvy

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    Relative to the person on the skate board the initial angle is, indeed, 90 degrees. The time the ball takes to return to the person's hand can be calculated entirely in terms of the vertical speed and acceleration.
     
  8. Jun 3, 2006 #7
    0 = 4t + 1/2 x -9.8 x t^2
    0 = 4t - 4.9t^2
    4.9t^2 = 4t
    t = 4/4.9
    t = 0.816 secs
     
  9. Jun 3, 2006 #8

    Hootenanny

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    Spot on. So you've now answered question (b). For question (c), all you need to find now is how far the ball would travel horizontally with a velocity of 5 m.s-1 in 0.816 seconds.

    ~H
     
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