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Relative projectile motion

  1. Jul 31, 2006 #1
    Hello,

    This is my first post here and I have the following question, please:

    A body is thrown vertically upward from the surface of the earth with an initial velocity of 7 m/s. Simultaneously an elevator cab starts to move upward from the same height the body was thrown from (the surface of the earth) with a constant velocity of 3 m/s. What is the max. height the body reaches relative to the cab's floor?

    Two possible solutions I've thought of:

    1- I calculated the max. height the body reaches relative to earth's surface which was 2.45 m, and it took the body 0.7 sec to reach its max. height, and then I subtracted the height of the cab after 0.7 sec (= 2.1 m) from the 2.45 m max height and got 0.35 m as the max height of the body relative to the cab's floor.

    or 2- I cosidered the cab as a frame of reference as it moves with a contant velocity, and that the body was thrown vertically with an initial velocity of 4 m/s (relative to the cab) and then used the equation:
    V^2=Vi^2 + 2*g*H-max
    with V=0 at max height and Vi=4 m/s (V of the body relative to the cab) and g=-10 m/s^2 we get:

    H-max= max height = 16/20 = 0.8 m

    Now, which solution is the right one (1 or 2 - 0.35 m or 0.8 m) and why?

    Thanks in advance!
     
    Last edited: Jul 31, 2006
  2. jcsd
  3. Jul 31, 2006 #2
    The second one is correct. Becouse in the first possiblety you falsely assume thet the maximum distance from the ground and the maximum distance from the elevatore are reached simultaneously, when in fact the maximum distance from the elevator is reached sooner.
     
  4. Jul 31, 2006 #3
    Thanks for your reply LENIN!
    Any other opinions, please?

    Cause my teacher insists that the first solution is the right one, and I really need a strong argument (no relativity stuff please - all classical) to prove him wrong!
     
  5. Jul 31, 2006 #4
    Well - considering Newton's classical equations of motion - the maximum height from the ground and the maximum height from the elevator are indeed reached simultaneously! AFAIK!

    Any other opinions!
     
  6. Jul 31, 2006 #5
    No they are not. When it's at the meximum distance from the elevator it's speed is 3 m/s upwards (relative to the ground) and when it's at maximum distance from the ground the speed is 0 m/s. As you can see the events can't be simultaneous.

    As for the argument I belive that your argument about the cab as a frame of reference should suffice.
     
  7. Jul 31, 2006 #6

    Office_Shredder

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    The second one is correct. I suggest you actually calculate how long it takes to reach the events described in the second solution, then show that, using the earth as the frame of reference, they actually are separated by .8 m

    Then you'll realize that it MUST be the correct answer, because .8 m is clearly greater than .35 m

    Another way to look at it... when the ball is thrown up, it decellerates. However, at first it still travels faster than the elevator. Once the ball reaches a speed of 4 m/s, it travels slower than the cab. It is at this point that, instead of gaining distance on the cab, the cab begins to catch up to the ball, thus cutting down on the distance
     
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