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Brain cramp on 2nd elevator problem

  1. Jun 9, 2004 #1
    "An elevator is moving upward at a constant velocity of 2.5 m/s. A ball is launched straight upward from the floor of the elevator with an initial velocity of 4.9 m/s (relative to the floor). (1) Determine the time the ball takes to reach its highest point from the standpoint of a person not moving outside the elevator. (2) Determine the time the ball takes to reach its highest point from the standpoint of a person in the elevator."

    OK, so in a normal (non-elevator) situation, the time it takes for a projectile to reach it highest point is just vi/9.8, which here is 0.5 sec.
    To me, the answer to both questions is 0.5 sec. Am i missing something here?

    Regards
     
  2. jcsd
  3. Jun 9, 2004 #2

    arildno

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    Wherever the ball is placed locally within the elevator, surely the highest point it will reach (relative to the observer outside), is when the elevator itself has reached the top floor :biggrin:
     
  4. Jun 9, 2004 #3

    NateTG

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    What's the initial velocity of the ball w.r.t. someone who is not on the elevator?

    Alternatively consider the following:

    If the ball is moving down at 1 m/s w.r.t. to the elevator it has passed it's highest point w.r.t. the elevator, but it's still going up at 1.5 m/s relative to the floor, so it has not passed it's highes poin w.r.t. the floor. Therefore the two times must be different.
     
  5. Jun 9, 2004 #4

    arildno

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    Since the ball was launched from the FLOOR of the elevator, the ball will never leave the elevator (unless there is an opening in the ROOF in the elevator).
    Hence, for a stationary observer outside the elevator, the highest point the ball will reach, will in all probability be at some time when the elevator has reached the top floor (of the building it's in).

    I think, however that the intended answer is along NateTG's lines...
     
  6. Jun 10, 2004 #5
    I guess I misunderstood the meaning of "highest". Thanks
     
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