Answer :
The number of floors from the 12th floor that the ball should descend in 3s is 9 when a ball is released from rest from the twentieth floor of a building.
From Newtons laws of motion we know that S = ut + 1/2at^2
where u denotes initial velocity, t denotes time taken, a denotes acceleration of the body and s denotes distance.
Given t = 1s, u = 0m/s since at rest and a = g = 9.8m/s^2
Then S = 0x1 + 1/2x9.8x1x1 = 4.9m
So the height of first floor == 4.9m
Then after 3s the distance covered by ball is S = ut +1/2at^2
S = 0x3 + 1/2x9.8x3x3 = 44.1m
Hence the distance covered by ball after 3s is 44.1m
Then the number of floor covered = 44.1/4.9 = 9
Therefore, we can conclude that the 9th floor is the floor from which the ball should descend in 3s starting on the 12th floor.
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complete question: A ball is released from rest from the twentieth floor of a building. After 1 s, the ball has fallen one floor such that it is directly outside the nineteenth-floor window. The floors are evenly spaced. Assume air resistance is negligible. What is the number of floors the ball would fall in 3s after it is released from the twentieth floor?
a. 3 floors or less
b. 4 to 6 floors
c. 7 to 10 floors
d. 11 floors or more