Problem 2. (EXTRA CREDIT). You are riding up in the elevator of the Cathedral of
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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 object 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)?Explanation / Answer
We need to analyse this using theory of Relativity. First, we will calculate the values with respect to Elevator (moving at constant speed of 2m/s.
We know that the coin goes up to 1 m in air above hand. Using v2 =u2 + 2*a*s, we can calculate the initial speed of coin w.r.t. elevator; here v = 0m/s at peak ; s =1m & a = -g (acceleration due to gravity)
So,initial speed of coin w.r.t. elevator, ue = SQRT(2*g*s) = SQRT(2*9.8*1) = 4.427 m/s
Now, time to reach peak can be calculated using v = u + a*t, where v= 0, a = -g
Time to reach peak w.r.t. person in elevator; tep = (ue) /g = 4.427 / 9.8 = 0.4517 second
Similarly, time taken to come down the coin from peak to hand (teh) can be calculated by
s =u*t +(1/2)*a*t2 ; here, s- 1m, u =0 (speed at peak); a = g ; t = teh
So, 1= 0*teh + (1/2)*g*teh2
teh = SQRT ( 2/ 9.8) = 0.4517 second
Now, we can analyse the coin movement with respect to Ground. Before throw of coin, speed of coin will already be speed of elevator (2m/s).
At the time of throw of coin, initial speed of coin w.r.t ground = ue + 2 = 4.427 + 2 = 6.427 m/s
Now, we can calculate the Peak of coin w.r.t. ground(sgp) using v2 =u2 + 2*a*s; where v = (at peak w.r.t. ground), u= 6.427m/s, a = -g = -9.8m/s2
02 = 6.4272 - 2* 9.8 * sgp
sgp = 2.107 m
Time to reach the coin to peak w.r.t. ground (tgp)can be calculated using v = u + a*t, where v= 0 at peak), u = 6.427m/s, a = -g = -9.8 m/s2
0 = 6.427 - 9.8 * tgp
tgp = 6.427 / 9.8 = 0.6558 second
As person on ground and person in elevator will see same time of the coin motion in air ; i.e. total time when coin was free in air.
Time taken to reach coin to hand from peak w.r.t. can be calculated as
tgh = (tep + teh) - tgp = 0.4517 + 0.4517 - 0.6558 = 0.2477 second
W.r.t ground, distance travelled from Peak to hand,
sgh = 0*tgh + (1/2) *g *tgh2 = 0.5*9.8* 0.24772 = 0.300 m
Based on above data, we can answer part a & b as follows:
Difference in time when coin reaches its peak w.r.t elevator and w.r.t. ground is
tgp - tep = 0.6558 - 0.4517 = 0.2041 second (anwer for part a)
Person on ground see the coin move from hand to peak w.r.t. ground = sgp = 2.107 m
Person on ground see the coin move from peak to hand w.r.t. ground = sgh = 0.300 m
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