There is a myth that dropping a coin from the Empire State Building can cause it
ID: 1552355 • Letter: T
Question
There is a myth that dropping a coin from the Empire State Building can cause it to become a deadly projectile by the time it reaches the ground, we will be looking at some of the physics behind this myth. Consider the situation is: The Empire State building has a height of 381 metres, while the mass of the coin is 3.11 grams.
Exercise1:
Using the information given, find:
- The time taken for the coin to drop to the ground from the top of the Empire State Building.
- The velocity at the moment of impact. * Assume a vacuum (no drag/air resistance).
Reflection1:
- Assuming the coin has a diameter of 19.05 mm and takes 10 ms to stop after hitting the ground, find the force and pressure that would have been exerted on the ground.
- If it takes approximately 689,476 Pa (~689.5 kPa) to break skin, will this coin penetrate a person’s skin?
Exercise2:
In reality there is drag/air resistance in the system which causes the coin to reach terminal velocity.
From experimentation the terminal velocity of the coin is found to be 50 km/h.
- Approximately how long would it take for the coin to reach this velocity?
- At approximately what height from the ground would this happen?
Reflection2:
- Re-do the reflection task from exercise 1 but take into account only terminal velocity was reached.
What was the force and pressure exerted on the ground in this case?
- If it takes approximately 689,476 Pa (~689.5 kPa) to break skin, will this coin penetrate a person’s skin?
Explanation / Answer
Given
height of the Empire State Building is h = 381 m
mass of coil is m = 3.11 g = 3.11*10^-3 kg
when the coin dropped from the top of the building then the gravitational potentila energy of coin turns in to k.e when it hits the ground
mgh = 0.5*mv^2
the velocity of the coin before impact is v = sqrt(2gh)
exercise --1
time taken to reach ground is s= ut+ 0.5 gt^2
h = 0 + 0.5 gt^2
t = sqrt(2h/g)
t = sqrt(2*381/9.8) s
t = 8.82 s
the velocity the moment of impact is v = sqrt(2gh)
V = sqrt(2*9.8*381) m/s
V = 86.42 m/s
Reflection 1
diameter of the coin d = 19.05 mm , radius is r = 9.525*10^-3 m
area of the coin is A = pi*r^2 = pi*(9.525*10^-3)^2 m2 = 0.0002850 m2
time taken is t = 10 ms
the impulse is j = F*t = m(v2-v1)
==> F = m(v2-0)/t
F = 3.11*10^-3(86.42)/0.01 N= 26.87662 N
the pressure is P = F/A = 26.87662/0.0002850 Pa = 94303.929824 Pa = 94.304 kPa
if the pressure required to break the skin is 689.5 k a then in this case , the coin will not penetrate the skin.
Exercise 2
if terminal velocity is v = 50 kmph = 13.89 m/s
v = v = sqrt(2gh)
13.89 = sqrt(2*9.8*h)
h = 9.8435 m
time taken to reach the ground is t = sqrt(2h/g)
t = sqrt(2*9.8435/9.8) s
t = 1.42 s
Reflection 2
diameter of the coin d = 19.05 mm , radius is r = 9.525*10^-3 m
area of the coin is A = pi*r^2 = pi*(9.525*10^-3)^2 m2 = 0.0002850 m2
time taken is t = 1.42 s
the impulse is j = F*t = m(v2-v1)
==> F = m(v2-0)/t
F = 3.11*10^-3(86.42)/1.42 N= 0.18927197 N
the pressure is P = F/A = 0.18927197/0.0002850 Pa = 664.11218 Pa
the coin will not penetrate the skin
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