If you dropped a cement block out of a plane at 5,ooo ft. would it pick up speed the entire time it was falling?
It depends on the aerodynamic drag. Since the aerodynamic drag is probably less than a human it would fall faster than 120 mph but theoretically it will reach terminal velocity at some speed. That is about all you can say with certainty. The question is: Will it reach that speed before it reaches the ground? D = (1/2)gT^2 5000ft = .5 * 32 f/sec/sec * t^2 t^2 = 5000ft/(.5*32ft/sec/sec) t^2 = 312.5 sec^2 t = 17.7 seconds After 17.7 seconds the velocity would be: v = gt = (32ft/sec/sec)(17.7sec) = 566 ft/sec 566ft/sec * (1 mile/5280 ft)* (3600 seconds/hr) = 385 mph If the wind resistance of the brick is 1/3 of a man I would guess that it reaches the ground before it reaches terminal velocity. By the way, this is the concept behind inertial bombs. You don’t need explosives. That’s what they used to shoot down the satellite. The force of the brick hitting the ground would be: Changing units and assuming a 15kg brick at 500 m/sec .1 second deceleration and other things: F = ma = m(dv/dt) = 1
