How fast would a penny, dropped from the 102nd floor of the Empire State Building, be traveling immediately before it hits the ground?

looney_tunes

The building's shape means a penny dropped from anywhere near the top would not hit the ground, but would land somewhere on the building. Pretty near the top, in fact. You would have to throw it horizontally with a great deal of force to produce a trajectory that would reach the ground. http://www.google.com.au/imgres?imgurl=http://www.empirestatebuildingnewyork.com/wp-content/uploads/2010/04/11.jpg&imgrefurl=http://www.empirestatebuildingnewyork.com/tag/empire-state-building-facts/&h=800&w=640&sz=43&tbnid=fU6UDBSD-rYGdM:&tbnh=143&tbnw=114&prev=/search%3Fq%3Dempire%2Bstate%2Bbuilding%26tbm%3Disch%26tbo%3Du&zoom=1&q=empire+state+building&hl=en&usg=__7RP08IYxTXQwPGqtDxQu-nK0Igc=&sa=X&ei=EBfcTd7gLIWgvQPqgsinDw&ved=0CGUQ9QEwBA&biw=1366&bih=578 If you wish to stipulate the necessary horizontal projection, then the coin would almost certainly have reached terminal velocity by the time it reached the ground. Its value would depend on how the coin was orientated during its flight, but could be close to 320 km/h (200 mph or 90 m/s), terminal velocity for a bullet. http://en.wikipedia.org/wiki/Terminal_velocity May 24 11, 2:43 PM

Baloo55th

"Its value would depend on how the coin was orientated during its flight" - surely it would still have a value of one cent whichever way up it was? :-) May 24 11, 2:53 PM

looney_tunes

Poorly worded in a rush to get off to work on time; I was referring to the value of the terminal velocity, not that of the coin. (It always looks strange to see times like 2:43pm for posts made at 6:43am my time. :D) May 25 11, 12:04 AM

navonil1

We know, H= u*t + (g.t^2)/2 Now, u=initial velocity=0 g=9.8m/s^2 H=1224 feet or 373m. therefore, t=((2*H)/g)^0.5 =((2*373)/9.8)^0.5 =8.7 seconds Hope it helps. Height known from: http://en.wikipedia.org/wiki/Empire_state_building May 25 11, 4:00 AM

navonil1

Now, if you want to know terminal velocity, v = (u^2 + 2*g*H)^0.5 = (2*9.8*373)^0.5 = 85.5 m/s Data: http://en.wikipedia.org/wiki/Empire_state_building May 25 11, 4:05 AM

looney_tunes

That formula for terminal velocity is quite simplistic. It is the final velocity that would be achieved if there were no air resistance as the coin fell that stipulated distance. But there will be air resistance, especially since the coin has to have a horizontal component to its motion. Terminal velocity is the fastest velocity that a particular object can attain as it falls through the resisting medium of the air. As it falls faster, air resistance increases so that its acceleration is no longer 9.8m/s/s. At some speed which depends on the object's shape acceleration ceases and it reaches terminal velocity. Nevertheless, that answer is close to the predicted terminal velocity of about 90 m/s. (You use H as 373m, but your link gives it as 381m. that changes your final velocity calculation upwards slightly.) May 25 11, 5:47 AM

gmackematix

According to the special effects experts Adam Savage and Jamie Hyneman in the 2003 series of the Discovery Channel's "Mythbusters", the upper limit of the terminal velocity of a penny is about 64 mph, not enough to do much damage to either concrete or a human head. http://www.youtube.com/watch?v=PHxvMLoKRWg May 25 11, 4:20 PM