How long does it takes and object at reat to drop 72'?

Question

can u please explaine how u get ther answer

Answer 1

s=(1/2)at² where s in distance (in meters) a is acceleration (in m/s²) and t is time (in seconds). The Earths gravitational field has a surface acceleration of 9.8 m/s² so
√(2s/a) = t
72" is 1.8288 meters and
√(2*1.8288/9.8) = about .611 sec.

Doug

Answer 2

Here's a formula you probably already know:

average speed = distance / time

Using variables:
v_avg = d/t

When something falls from rest for "t" seconds, its FINAL speed is (g)(t), where g = acceleration due to gravity = 32 ft/sec²

The AVERAGE speed of something falling from rest, is just 1/2 of its final speed. So:

v_avg = v_final/2 = (g)(t)/2

equating that with the earlier expression for v_avg gives:

d/t = (g)(t)/2

Solving for t:
t = sqrt(2d/g)

plugging in known values:
t = sqrt(2(72')/(32 ft/sec²)) = 2.1 sec.