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Projectile Motion : The Impossible Problem

difficulty

Poll closed Jun 11, 2009.
  1. easy

    0 vote(s)
    0.0%
  2. medium

    4 vote(s)
    100.0%
  3. hard

    0 vote(s)
    0.0%
  4. impossible

    0 vote(s)
    0.0%
  1. Jun 1, 2009 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown straight up from the edge of the roof of a building. a second ball is dropped from the roof 2 seconds later.

    a) if the height of the building is 6.00 m, what must be the initial speed of the first ball if both are to hit the ground at the same time?

    consider the same situation but now let the initial speed of the first ball be given and treat the height of the building as an unknown

    b) what must be the height of the building be for both balls to reach the ground at the same time for each of the following values of initial velocity:
    i) 13.0 m/s
    ii) 19.2 m/s

    c) if initial velocity is greater than some value vmax, there is no value of h for which both balls hit the ground at the same time. solve for vmax. the value of vmax has a simple physical interpretation. what is it?

    d) same question as part c except we are looking for an initial velocity some vmin below which there is no value of h for which both balls hit the ground at the same time. solve for this vmin and give the physical interpretation


    2. Relevant equations

    vf2 = v02 + 2a (y-y0)

    average acceleration = change in velocity / change in time
    average velocity = change in position / change in time

    y = y0+vy0t - (1/2)gt2

    vf = v0 +at

    x-x0= ((v-v0)/2)t


    3. The attempt at a solution

    clues from the professor: find a general expression combining some basic projectile formulas relating height of the building and initial velocity somehow using times t and t+2 seconds to combine some of these equations, to solve for a, then manipulate to solve for b, and manipulate again to solve for c and d. Sorry guys I thought i had solved a but I think made an error in logic, so posting what i have so far would probably confuse you further.
     
  2. jcsd
  3. Jun 1, 2009 #2

    mgb_phys

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    Re: Projectile Motion

    For a the speed that it comes back down to the edge of the building is the same speed it was thrown up at (ignoring air resistance).
    So you have a ball dropped and 2sec later one thrown down - just play with the equations of motion
     
  4. Jun 2, 2009 #3

    Hootenanny

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    Okay, so let's start by writing an expression for the position of the first ball at time t. Can you do that?

    Edit: Threads merged.
     
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