Imagine riding up in a glass-walled elevator. It’s a long ride. To pass the time
ID: 1413145 • Letter: I
Question
Imagine riding up in a glass-walled elevator. It’s a long ride. To pass the time you toss and catch a ball, straight up and down, about 1 meter into the air above your hand. You watch the ball go up, reach the peak of its toss, and come back down into your hand, and it occurs to you to wonder: If people are watching the ball’s motion from outside of the elevator, what do they see it doing? (a) Would they see the ball reach the peak of its toss at the same instant you see the ball reach the peak of its toss? If not, figure out what the difference in time is, between the instant they see the ball reach the peak of its toss and the instant you do. Explain in plain and simple terms why your answer makes sense. (b) You see the ball move 1 meter up and fall that same meter back down. How far up do they see the ball move (from when you release it to when they see it reach the peak) and how far down (from the peak to when you catch it)?
Explanation / Answer
The elevator is moving up with a certain speed, since the velocities are very small we can use the galilean assumptions which means that the time when the events happen is the same for both observers, meaning that they see the ball reach the peak of its toss at the same instant you see the ball reach the peak of its toss
But the distance traveled is different because you are on an elevator at the same speed that the ball, the person outside will see more distance traveled because he/she is stationary. The amount of distance would depend on the elevator speed which is not provided but it would be larger if seen from the observer's reference frame.
The velocity of the ball can be calculated using energy conservation
mgh=mv^2/2
if h=1 m g=9.8
v=4.42 m/s
The ball's velocity of the observer's reference frame is v'=v_elevator+4.42
so the distance will depend on the elevator's velocity also