Feb 11, 2009 #1 D DANIELM Member Joined May 18, 2008 Messages 156 Reaction score 0 Points 16 ...hit the ground? One of our operatives fell roughly this distance onto concrete and very fortunately got up only winded. I don't know how to calculate this, help appreciated.
...hit the ground? One of our operatives fell roughly this distance onto concrete and very fortunately got up only winded. I don't know how to calculate this, help appreciated.
Feb 11, 2009 #2 L Let'slearntothink New member Joined Jan 23, 2009 Messages 8 Reaction score 0 Points 1 v^2 = 2gh =2x32x20 = 1280 0r v = 35.78 ft/s. The man will travel with 35.78 ft/s velocity when he hits ground.
v^2 = 2gh =2x32x20 = 1280 0r v = 35.78 ft/s. The man will travel with 35.78 ft/s velocity when he hits ground.
Feb 11, 2009 #3 L Let'slearntothink New member Joined Jan 23, 2009 Messages 8 Reaction score 0 Points 1 v^2 = 2gh =2x32x20 = 1280 0r v = 35.78 ft/s. The man will travel with 35.78 ft/s velocity when he hits ground.
v^2 = 2gh =2x32x20 = 1280 0r v = 35.78 ft/s. The man will travel with 35.78 ft/s velocity when he hits ground.