ball problems

Discussion in 'Physics' started by dileepchacko, Jul 31, 2008.

  1. dileepchacko

    Thread Starter Active Member

    May 13, 2008
    102
    1
    Hi All

    A man is standing on the top of the building, 30m height. He drops a ball from the top, at a same time another man standing on the ground floor, throwing a ball at a initial speed of 30m/s. In what time two balls will collide each other.
     
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    May we assume the balls are initially traveling toward each other? That the man on the ground threw the ball up whilst being directly under the man on the roof?

    May we further assume the balls are sufficiently dense as to disregard drag?
     
  3. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    dileepchacko,

    Not too hard really. First some conventions. Up is defined as the positive direction. Therefore gravity accelerates in a negative direction.

    Any good physics textbook will give the formula for distance vs time in a constant acceleration. I can derive it if need be. It is:

    s = 1/2(G*T^2) + Vo*t + So

    Vo = 30 m/s for man on the ground, 0 m/s for man on building

    So = 30 m for man on building, 0 m for man on ground

    Equating the positions of the two balls:

    -1/2(G*T^2) + 30 = -1/2(G*T^2) +30*T

    Which is easily solved to T = 1 sec.

    Ratch
     
    Last edited: Jul 31, 2008
  4. dileepchacko

    Thread Starter Active Member

    May 13, 2008
    102
    1
    Answer for the ball problem.


    Here the logic is simple, a freely falling body in the earth will be having acceleration of 9.8m/s. The initial velocity of the ball which the man drops from the top of the building will be zero, but the initial velocity of the ball which is through from the ground floor will be 30m/s. The balls are moving opposite direction. Equate the poison of both balls, because both balls are colliding each other after some time.


    From the newtons law of motion, three equations can be derived.


    V = U + AT -------------------------------------- (1)


    S = UT + (½) AT2 + S0. ---------------------------------------(2)


    V2 = U2 + 2AS + ---------------------------------------(3)


    where.


    V = Velocity of an object.
    U = Initial velocity of an object.
    S = Distance traveled by the object.
    T = Time period of the object.
    A = Acceleration of the object.
    S0 = initial distance


    Take the first case, which man drops the ball


    U = 0
    A = 9.8m/s substitute these result in equation (2)


    S = 30 + (½) * 9.8 * T2 ------------------(4)




    Take the second case, which man throws the ball


    S = 30T + (½) * 9.8 * T2 -------------------(5)


    equate (4) and (5)


    we get 30T + (½) * 9.8 * T2 = 30 + (½) * 9.8 * T2


    T = 1 second
     
  5. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    dileepchacko,

    Observe the following points:

    1) For this problem, it doesn't matter what the strength of the gravity field is. It will still take 1 sec for collision on the moon, Mars, Jupiter, or no gravity at all.

    2) In text, V2 does not mean V squared. Use V^2 to denote exponentiation.

    3) The equations you designated (1) and (3) are irrelevant to the problem solution. By the way, can you derive those three equations?

    Why did you bother to post the solution when I already did?

    Ratch
     
  6. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Note also the building height and lower ball launch velocity are not vital, so long as they are of equal magnitude. In other words, one building height per second for the lower ball regardless of building height. Meters, miles, astronomical units, doesn't much matter until we approach light-seconds.
     
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