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Projectile Motion including Displacement and Speed

  1. Aug 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Q. Jessica and Tammy are playing a game of tennis on a court that is 24m long.

    Tammy is standing at the net, so Jessica (who is at the baseline) lobs the ball from near the ground aiming for the baseline on Tammy's side 24m away. She hits the ball at an angle of 60◦ above the horizontal,

    a) If the ball lands on the baseline 24m away, what would the displacement of the ball be?
    b) How fast was the ball going just after Jessica hits the lob?

    Tammy manages to run back, and using a backhand shot, returns the ball from the baseline just as it is about to contact the ground. The ball just skims the 0.90m high net at the top of its flight, landing 24m away,

    c) What would the total displacement of the ball be when it lands back at Jessica's baseline?
    d) And how fast is the ball travelling when it leaves Tammy's raquet?



    2. Relevant equations

    Vector Components
    Vxo = Vo COS (theta)
    Vyo = Vo SIN(theta)

    Displacement
    X = Xo + Vxot + 0.5 gxt^2

    3. The attempt at a solution

    a)
    I've tried to work out the individual vector components:
    for x:
    Vxo = Vo COS (theta)
    Vxo = 24 Cos(60◦)
    = 12 m


    From there I thought I could use the equation to work out the displacement for x, as we are dealing with a horizontal displacement (right??), as the ball is on the baseline, so y=0 in this case

    So, displacement, where g = 9.8
    X = Xo + Vxot + 0.5 gxt^2
    X = 0 + 12m + 0.5 gxt^2

    I know I have to do some form of quadratic manpulation here.. which i'm not entirely sure of.. as I don't have t.

    Also, i'm not sure if i'm going in entirely the wrong direction, cos from the question, it's asking for displacement at the baseline, so I was almost fooled into thinking that it was a scalar calculation, which would equal the length of the court, i.e. 24m.. but i think that's wrong too!

    So, i'm stuck on the easiest part a) for now... I haven't gotten past that to look at the other parts yet.. hopefully i'll see the light soon!

    Thanks for any help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 8, 2008 #2

    Kurdt

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    Gold Member

    The displacement of the ball is just how far away it is from where it started. Since it lands 24m away, that will be the displacement.
     
  4. Aug 8, 2008 #3
    Ahhh.. ok! God I feel stupid, I knew I was over-complicating things.

    Thanks for replying Kurdt.

    So, that would mean for part c that when the ball is returned, the displacement will be 0, using the same principle of part a?

    Also, can someone please clarify for me some terminology for when i'm studying up on this... is there a difference between the term "Range" which i'm just learning about, and displacement (horizontal) - so is it valid to answer that the Range in this is example is 24m for part a, for example?

    Thanks!
     
  5. Aug 8, 2008 #4

    Kurdt

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    Science Advisor
    Gold Member

    Yes the displacement will be zero in part c. Range is normally the distance you want a projectile to travel while displacement is the difference between the final position and starting position.
     
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