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?

Answer 1

Answer:

9 floors

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 1+\frac{1}{2}\times 9.81\times 1^2\\\Rightarrow s=4.905\ m$

So, one floor is 4.905 m tall

In 3 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 3+\frac{1}{2}\times 9.81\times 3^2\\\Rightarrow s=44.145\ m$

The distance the ball would have covered would be 44.145 m in 3 seconds

Number of floors

$\frac{44.145}{4.905}=9$

So, the ball would have fallen 9 floors in 3 seconds