Question

Problem 2. (EXTRA CREDIT). You are riding up in the elevator of the Cathedral of Learning, tossing a coin 1 meter into the air above your hand. You watch it go up, reach the peak of its toss, and come back down into your hand, and it occurs to you to wonder: If people could watch its motion from outside of the elevator, what would they see it doing? Of course, they also would see that the elevator is moving, at a rate of about 2 m/s. But suppose they just pay attention to the coin a. Consider that peak means the farthest away an objeet thrown in the air gets from the observer. Would a person on the ground see the coin reach the peak of its toss at the same instant you see the coin reach its peak? If not, calculate the difference in time between the instant they see the coin reach the peak of its toss and the instant you do. b. You see the coin travel 1 meter up and fall the same meter back down. How far up does the person on the ground see the coin 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)?

Answer 1

We need to analuse this using theory of relativity first, we calculate the values with respect to Elevator (moving at constant speed of 2 m/s) We know that the coin upto 1 m in air above hand using v^2 = u^2 + 2 as, we can calculate the initial speed of cion write elevator here v = 0 m/s at peak s = 1 m and a = -g (acceleration due to gravity) So, initial speed of coin write elevator, u_e = squareroot 2 gs = squareroot 2 times 9.8 times 1 = 4.427 m/s Now, thime to reach peak can be calculate using V = u + at Where V = 0, a = -g Time to reach peak write person in elevator t_ep = (u_e)/g = 4.427/908 = 0.4517 second Similarly time taken to come down the cion from peak to hand can be calculate by S = ut + 1/2 at^2 Here S = 1 m u = 0 (speedat peak) a = r t = ten