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Ball caught by player, solve quadratic equation

  1. Oct 17, 2014 #1
    1. The problem statement, all variables and given/known data

    First part of the question (SOLVED)

    A football player kicks a football so that the angle of incidence is 50 DEG and the initial magnitude of velocity of the ball is 15 m/s.

    Find the:
    a) Balls maximum height = 6.7 m
    b) Time of flight = 2.3 s
    c) time when the ball reaches the maximum height = 1.2 s
    d) horizontal distance = 23 m
    e) velocity at impact = 15 m/s [50 DEG below horizontal]

    Second part of the question (Need help)

    Now imagine that the situation is exactly the same, except this time, the ball is caught by a player when it is at a height of 1.5 m above the ground. Determine the horizontal range of the ball, the length of time that it was in the air, and the final velocity.

    2. The attempt at a solution

    Let [up] be the positive y-direction and [forward] be the positive x-direction.

    Vertically
    V1y = +11.49 m.s
    Ay = -9.8 m/s
    Total y displacement = 1.5m

    Using the vertical component to find time:

    Displacement y = V1y (change in time) + 1/2 Ay (change in time)^2
    1.5 m = (+11.49 m/s)(change in time) + 1/2 (-9.8 m/s^2)(change in time)^2

    There is a quadratic equation to solve in this equation so it has been rearranged

    (9.8 m/s^2)(change in time)^2 + (-11.49 m/s)(change in time) + 1.5m = 0

    Now my workbook gives the rearrangement differently but doesn't explain why and this is where I need help.

    My workbook says the rearrangement should look like this:

    (4.9 m/s^2)(change in time)^2 + (-11.49 m/s)(change in time) + 1.5m = 0

    Can someone please explain why the acceleration was divided in two?
     
  2. jcsd
  3. Oct 17, 2014 #2

    Doc Al

    User Avatar

    Staff: Mentor

    See the factor of 1/2? That's what you lost.

    That factor of 1/2 is from the kinematic formula for displacement. You must have dropped it and not noticed when you did your rearrangement.
     
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