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Catching a thrown ball

  1. Jul 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Starting from rest, a child throws a ball of mass m with an initial speed v , at an angle B with the horizontal direction. The child then chases after the ball, accelerating at a constant acceleration a . If the child wants to catch the ball at the same height as it was thrown, what must be the child's acceleration a ? Express your answer in terms of some or all of the variables v , m , B and g for the gravitational constant. Express the trigonometric functions in terms of the basic sin(B), cos(B) or tan(B).

    2. Relevant equations
    1-D Kinematics

    3. The attempt at a solution
    I'm using the point of release as the origin, positive-y is up and positive-x is the direction of the throw. I'm interested in the situation where at time t_f, x_child(t) = x_ball(t) and y_ball(t)=0. I start with the vertical.
    t_f is the moment of the catch
    v_yi is the initial vertical component of the ball's velocity; v_xi is the initial horizontal component of the ball's velocity

    y(t)=v_yi*t-1/2*g*t^2
    t_f=2*v_yi/g

    For the horizontal component:
    x_ball = v_xi*t
    x_child = 1/2*a*t^2
    At t_f:
    v_xi*t_f=1/2*a*t_f^2
    Substituting gives:
    a=g*v_xi/v_yi

    With vector decomposition I get an answer of:
    a = g*v*cos(B)/sin(B)


    What am I doing wrong here?
     
  2. jcsd
  3. Jul 19, 2015 #2

    TSny

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    Homework Helper
    Gold Member

    Your work is very nicely written out :smile: .

    Should the factor of v be in your final answer? Doesn't v cancel out?
     
  4. Jul 19, 2015 #3

    Thanks, that does indeed solve it. Perhaps one day I'll learn to do basic algebra.
     
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