A penny is dropped from the observation deck of the Empire State building (370 m above ground). With what velocity does it strike the ground? Ignore air resistance.
distance = 1/2 gt^2 Solve for time it takes for the penny to reach the ground.
Then distance = 1/2(Vinitial + Vfinal)t.
Vinitial is zero. You know distance and t, solve for Vfinal.
12
85.2
d=1/2gt^2
t^2=2d/g
t^2=2(370)/9.8
t^2=75.51
t=8.6 second
v2=v1+gt
v2=0+(9.8m/s^2)(8.6s)
v2=84.3m/s
Well, let me do some quick calculations here... Okay, here's the answer: the penny strikes the ground with the velocity of "ouch, that's gonna hurt!" But in all seriousness, the velocity it would strike the ground can be calculated using the principles of free fall. By using the equation v = √(2gh), we find that the velocity is approximately 86 m/s. So, it definitely won't be a gentle landing!
To determine the velocity at which the penny strikes the ground, you can use the principles of classical mechanics.
The first step is to find the time it takes for the penny to fall. We can use the equation for free fall motion:
h = (1/2) * g * t^2
Where:
h is the height (370 m)
g is the acceleration due to gravity (9.8 m/s^2)
t is the time it takes to fall
Rearranging the equation to solve for t:
t^2 = (2 * h) / g
t = sqrt((2 * h) / g)
t = sqrt((2 * 370 m) / 9.8 m/s^2)
t ≈ 8.6 s
Now that we know the time of fall, we can find the velocity using the equation:
v = g * t
Substituting the known values:
v = 9.8 m/s^2 * 8.6 s ≈ 84.28 m/s
Therefore, the penny will strike the ground with a velocity of approximately 84.28 m/s.