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Homework Help: Ball falls from top of cliff; kinematics

  1. Sep 10, 2010 #1
    1. The problem statement, all variables and given/known data
    A ball is dropped from rest from the top of a cliff that is 17.4 m high. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is exactly the same as that with which the first ball eventually hits the ground. In the absence of air resistance, the motions of the balls are just the reverse of each other. Determine how far below the top of the cliff the balls cross paths.

    Ball 1:
    u1 (initial velocity)= 0 m/s
    g= -9.81 m/s2
    H (height of cliff)= -17.4m
    V1 (final velocity)= ?

    Ball 2:
    u2= V1= ?

    Unknowns:
    t= ?
    x (where balls meet above the ground)= ?


    2. Relevant equations
    V2=u2 + 2ax
    x = u(t) + 0.5(g)(t)2


    3. The attempt at a solution
    First, I calculated the final velocity (V1) of Ball 1 as so:
    V12=u2 + 2ax
    V1= sqrt[(0m/s)2 + 2(-9.81m/s2)(-17.4m)
    V1= 18.4673m/s ~18.47m/s

    Therefore, the u2 (initial velocity) of Ball 2 is also 18.47m/s.

    I know that the next step is to calculate the time by using x = u(t) + 0.5(g)(t)2, but I'm not sure on how to manipulate the equation to fit the criteria of this problem. Please help me!
     
  2. jcsd
  3. Sep 10, 2010 #2

    rl.bhat

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    Homework Helper

    Hi dansunefusee. welcome to PF.

    Both the balls start simultaneously. When they meet they must have traveled same duration t s. If x is the distance from the top of the cliff, then for the falling ball
    -x = -1/2*g*t^2..........(1)
    For the ball going up
    h - x = ut - 1/2*g*t^2 ........(2)
    Where h is the height of the cliff. Solve the two equations and find the time t.
     
  4. Sep 10, 2010 #3
    Thank you!! :) I'll need all the help I can get from PF this year. Haha.

    To make sure I am doing this right, I would need to combine the 2 equations just to find time?

    Given equations:
    1) -x = -1/2*g*t^2
    2) h - x = ut - 1/2*g*t^2

    Therefore, I would plug the 1st equation for the -x variable in the 2nd equation to get:

    h - (-1/2*g*t2) = ut - 1/2*g*t2

    Then isolate the time (t) variable by itself???

    :confused:
     
  5. Sep 10, 2010 #4
    Hey , after you take the time(t) , substitute it to (1) equation , then x will come,
     
  6. Sep 10, 2010 #5

    rl.bhat

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    Homework Helper

    This step is wrong.

    It should be h - 1/2*g*t2 = ut - 1/2*g*t2
     
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