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Kinematics long jumper problem

  1. Sep 10, 2008 #1
    Hi ...I came up another question again i guess you will help me out .Here is the question

    1. A long jumper takes off at an angle of 20 degree with the horizontal and reaches a maximum height of 0.55 m at mid-flight.
    - how long is she in the air?
    - what is the forward component of her velocity?
    - how far does she jump?

    Here is what i try.

    A. how long is she in the air.
    i use the formula Y= vt + 1/2gt^2
    which is -0.55 = 1/2 * -9.8 t^
    so t= 0.335 sec and finaly i multiplay 0.335 * 2 = 0.67 sec.
    B. here is where i stop how can i find her velocity with out knowing the horizontal distance.

    i know the formula X = vt

    thank you.
     
  2. jcsd
  3. Sep 10, 2008 #2

    LowlyPion

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    Re: problem

    Think about what the horizontal and vertical components are of her initial velocity.
    (Hint: They give you an angle.)

    Horizontal velocity has nothing to slow it so you are right that distance will be Vx * t.
    Vertical distance is given so you know the total time from doubling how long it would take for her to fall. And that you know as = 1/2 g*t2
     
  4. Sep 10, 2008 #3
    Re: problem

    yes i did but i still didn't get that because how can i find with only one given angle
     
  5. Sep 10, 2008 #4

    LowlyPion

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    Re: problem

    That's how you solve the problem.

    Vy = V*Sinθ is the Y component of the initial velocity.
    Vx = V*Cosθ is the horizontal component.
     
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