Free Falling Body (2)
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Free Falling Body (2)

(a) Given that v is the speed at time t of an object free falling to earth and d is the distance the object has fallen at time t. Derive a relation between v and d.
(b) Using the relation derived in (a), calculate the speed with which an object dropped from the top of a building 1000 ft above street level hits the ground.

The string: S7P5A51 (speed-physical change); S7P4A41 (Linear motion)
The Math:
Neglecting air resistance, an object falls to earth with a constant acceleration:
a = 32 ft/sec2.
A falling object with an initial velocity of zero, gains speed each second at the rate of 32 ft/sec.
So after t secs, speed v = 32t----(1).
avrage speed is attained after one-half of the time traveled.
So average speed = 32t/2 = 16t----(2)
So distance after t secs, d = average speed x time = 16t2---(3)
From equation (3), t = √(d/16)
substituting this value of t into equation (1), we have:
v = 8√d----(4)
Expressing equation (4) in terms of d, we have:
d = v2/64----(5)
To calculate speed of object when it hits the ground use equation (4)
So, v = 8√1000 = 256 ft/sec approximately.

Blessed are they that have not seen, and yet have believed. John 20:29