How Does the Elevator's Velocity Affect a Dropped Ball's Motion?

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The discussion revolves around a physics problem involving a ball dropped from an upward-moving elevator. The key points include the initial conditions of the ball, which should be considered with an initial velocity of 2 m/s due to the elevator's motion. The time it takes for the ball to hit the floor is calculated to be 0.78 seconds, as stated in the book. For the total distance traveled relative to the ground, the correct approach involves accounting for both the upward and downward motion of the ball, leading to a total distance of 1.84 meters. Understanding the frame of reference is crucial for solving such problems accurately.
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Hi, there's a question in the book that says:
"An elevator of height 3 m moves upward at 2m/s. A ball is dropped from the roof. a) when does it hit the floor? b) what is the total distance traveled relative to the ground?"

Now, the way I had set it up was that when the ball drops, it falls with an initial velocity of 0m/s and an acceleration of -9.8m/s^2. However, I also took into account the velocity of the elevator moving upwards at 2m/s... I was wondering why we disregard that velocity? (the answer in the book is 0.78s).

Also for part B, I'm not exactly sure how to reach the book's answer. What I did was use the total time, multiplied by the elevator's velocity, so 1.56m. Therefore, once the ball reaches the floor, 0.78 seconds would have passed and thus, it would be 1.56m from the ground. However, the answer is 1.84m, so I'm a little stumped. Any help is appreciated!
 
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Originally posted by Skyblitz
Now, the way I had set it up was that when the ball drops, it falls with an initial velocity of 0m/s and an acceleration of -9.8m/s^2. However, I also took into account the velocity of the elevator moving upwards at 2m/s... I was wondering why we disregard that velocity? (the answer in the book is 0.78s).
If the elevator moves uniformly, its motion will not affect any experiments done inside it. (See https://www.physicsforums.com/showthread.php?s=&threadid=10779)

If you want to include the speed of the elevator, be sure to do it completely. The initial speed of the ball with respect to the ground is +2 m/s, same as the elevator.
Also for part B, I'm not exactly sure how to reach the book's answer. What I did was use the total time, multiplied by the elevator's velocity, so 1.56m. Therefore, once the ball reaches the floor, 0.78 seconds would have passed and thus, it would be 1.56m from the ground. However, the answer is 1.84m, so I'm a little stumped. Any help is appreciated!
For part B, do everything with respect to the ground. You know the initial speed and height of the ball. It goes up, then down. You have to find the total distance.
 
Thanks a lot for your help!
 
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