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Accelerated Linear Motion

  1. Nov 2, 2003 #1
    Im having some trouble with this Accelerated Linear Motion question. I dont have any clue where i should even start with part (B).
    Im just lost, dont have a clue where to start. My head just starts to go into overload when i even think about it. Please help!
  2. jcsd
  3. Nov 2, 2003 #2
    Your formula

    V(final) = V (original) + acceleration * time


    Vf - Vo + at

    You have only 3 points but 4 velocities, so I assume, your original or initial velocity would be u.

    Have you tried this formula???

  4. Nov 3, 2003 #3
    Yes, ive used the formulae before, im having trouble with part two, the juggler question.
  5. Nov 3, 2003 #4


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    Staff Emeritus
    Science Advisor

    Part two is relatively simple.

    Each ball has acceleration -g= -9.81 m/s2. If we take "v" to be the initial velocity of each ball, then v(T)= v- 9.81T in m/s where T is measured in seconds since the ball was thrown.
    In addition, the height of each ball after T seconds is given by
    h(T)= vT- (9.81/2)T2.
    The ball will continue upward until its velocity is 0: v- 9.81T= 0 so
    T= v/9.81. We are told that the height at that time is 3 m:
    v(v/9.81)- (9.81/2)(v/9.81)2= v2/(2*9.81)= 3 so
    v2= 2*3*9.81 and v= &radi;(2*3*9.81).

    You can find t in either of two ways: directly- The ball comes back to the person's had (assumed same height) exactly t seconds after the sixth ball is throne up and there is and interval of t seconds between each one: the first ball comes back to the person's had after 6t seconds: h(6t)= v(6t)- (9.81/2)(6t)2= 0. You know v from the previous problem. Solve for t.
    Less directly: because of the symmetry of the motion, each ball attains it highest point exactly half way through its motion: 3t.
    Since you calculated that time, T, in the first problem, 3t= T. Solve for t.

    You know that when the first ball is again in the juggler's hand, time 6t has passed. Its height is0. The second ball is exactly "t" behind. Its height will be given by h(5t), the third ball's height will be h(4t), the fourth ball's height by h(3t), the fifth ball's height by h(2t), and the sixth ball's height by h(t). If the juggler throws the first ball up again the instant he catches it, each ball will have one of those 6 heights each time a ball is caught.
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