Approximately how long would it take for a 115-118 lbs object to fall about 10 ft?

The proper formula is s = Vi*t + (1/2)a*t^2 where s = about 10 ft, Vi = 0, and a = 32 ft/s^2

So t = sqrt(2*10/32) seconds, which is about 0.8 seconds.

Your other answer is the time it would take if the velocity was a constant 32 ft/s.

Not taking wind resistance into consideration, the same time that anything else would take to fall.

The acceleration rate for the earth is: 32.2ft/sec. So, an object dropped from ten feet will take .3105 seconds to hit the earth.

The answer depends on air resistance.

But assuming the object wasn't attached to a parachute we can probably ignore the effects of air resistance and use the simple formula:

t = √(2h/g)

Note this formula works for ANY weight (as long as air resisitance is negligible).

t = √(2 x 10 / 32)

. = √0.625

. = 0.79s

Wait, I have a computer program that'll work that for you:

drop ht = 3.05 meters

drop ht = 0.00 kilometers

drop ht = 10.01 feet

drop ht = 0.0019 miles

SUMMARY OF RESULTS

Object mass (M) = 50.0000 kg or a nominal 110.231 lbs

Coefficient of drag (Cd) = 0.400

X-section area (A) = 1.000000 meters^2

Starting altitude H = 10.01 feet or 3.05 meters

Starting speed U = 0.00 mps

Impact or Terminal V = 17.95 mph or 8.03 mps

Time to impact T = 0.790 seconds or 0.013 minutes

Maximum V = 17.95 mph at 0 feet

Altitude lost to reach Max V = 10.01 feet or 3.05 meters

.79 second

T = sqrt(2*10/32)