Ball in freefall

Posted on 2006-04-18
Last Modified: 2008-01-09
A ball of mass 1.5kg is thrown upward. It leaves the thrower's hand with a velocity of 10m/s. The following questions refer to the motion after the ball has lef tthe thrower's hand. Assume that the upward direction is positive.

a) How long does it take for the ball to return to the throwers hand?
I suppose I could do this problem using simple 1-dimensional kinematics....but i've been learning momentum, impulse etc. So i'm trying to do this problem using my new knowledge :/

The only thing I could think of that involved time was the fact that Impluse is the integral from [t1->t2] of f(x) dx, where f(x) is a force function. So i'm not really sure how to do this part.

b)  What is the final velocity of the ball (just before it reaches the hand).

   This is easy -v_1.
   v_2 = -v_1

c) What is the change in momentum of the ball.

   Again fairly easy.

   p_2 - p_1 = mv_2 - mv_1 = -mv_1 - mv_1 = -2mv_1 = -30 kg m/s

d) What is the impluse calculated from change in momentum.

   J = Change in P, thus...

   J = -2mv_1 = -30 N*s

e) What is the average force acting on the ball

   I think this might tie in to part (a), but i'm not sure.

Question by:BrianGEFF719
    LVL 84

    Accepted Solution

    a) in absense of friction, it will return with the same speed, so
    10m/s - -10m/s = at
    if a = 10m/s², t=2

    f*t = momentum change
    LVL 84

    Expert Comment

    or f=ma
    LVL 19

    Author Comment

    Thanks as always ozo.
    LVL 19

    Author Comment

    I get Average Force on the ball is 14.70N...

    Also, the question goes on to ask the following...

    After the ball hits the throwers hand it comes to rest in 0.25s

    What is the net implus exerted on the ball?

      J = change in p

      = -mv_1

      = -(1.5kg)(-10m/s) = 15 N*s

    What is the average force exerted by the hand on the ball (dont forget gravitational force)

     J = F_net * t

      J = (F_h + F_e)t
     J = F_h*t + mgt

     J - mgt
     --------- = F_h

    F_h = (15N*s + (1.5kg)(9.8m/s^2)(.25s))/.25s = 74.7 N

    Again doesnt seem resonable, is this work correct ozo?
    LVL 19

    Author Comment

    Because 74.7N * .25s is not equal to 15 N*s...i'm getting a number larger, which means I no longer have equality.
    LVL 19

    Author Comment

    >>Because 74.7N * .25s is not equal to 15 N*s...i'm getting a number larger, which means I no longer have equality.

    I retract that last statment, I solved for F_h, not F_net....i'm tired :/

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