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1 month ago

3

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 di

rectly 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?

1 answer:

serg [3.5K]1 month ago

4 0

Answer:

9 floors

Explanation:

Variables defined:

t = elapsed time

u = starting velocity

v = final velocity

s = vertical displacement

a = gravitational acceleration = 9.81 m/s²

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

Each floor has a height of 4.905 meters.

For a time of 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 ball will have fallen 44.145 meters in 3 seconds.

Number of floors:

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

Thus, the ball drops through 9 floors over 3 seconds.

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