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

Answer 1

The number of floors that the ball should fall in 3s from the 12th floor is the 9th floor.

We know that

$S = ut + \frac{1}{2at^2}$

where,

u denotes the initial velocity = 0 m/s.

t denotes the time = 1s

S denotes the displacement.

a denotes the acceleration = 9.8m/s^2

Now S should be

$= 0\times 1 + \frac{1}{2} \times 9.8 \times 1 \times 1\\\\= \frac{9.8}{2}$

= 4.9m

So, the 1 floor should be 4.9m tall.

Now After 3 seconds

$S = ut + \frac{1}{2at^2}$

$= 0\times 3 + \frac{1}{2} \times 9.8 \times 3^2\\\\= \frac{9\times 9.8}{2}$

= 44.1m

Now the number of floors should be

= 44.1 ÷ 4.9

= 9

Therefore we can conclude that the number of floors that the ball should fall in 3s from the 12th floor is the 9th floor.

Learn more: brainly.com/question/19930452