Question

1. A basket coffee filter (see below) is very light and has a large drag coefficient. It is possible to stack several filters together so they have the same drag coefficient as a single filter. Suppose you tried dropping one filter from a ladder, then tried dropping two stacked filters from the same height, then 3 and so on. For each try you measure the time taken for the filters to fall to the floor. How would you expect the time for the filters to fall to compare to the number of filters

Answer 1

he times are getting closer as we use each filter, in the expression n would be the number of filters

t =
`\sqrt{ (2(y_o - (n-1) h))/(g) }`

Step-by-step explanation:

For this exercise let's use the kinematics relations

y = y₀ + v₀ t - ½ g t²

When the first filter reaches the ground, its height is y = 0, as they release its initial velocity is zero

for the 1st filter

0 = y₀ - ½ g t²

t² = 2y₀ / g

t =
`\sqrt{ (2y_o)/(g) }`

when we release the second filter upon arrival it has a height y = h where h is the height of each filter

h = y₀ - ½ g t²

t =
`\sqrt{ (2(y_o- h))/(g) }`

when we release the third filter it reaches y = 2h

2h = y₀ - ½ g t²

t =
`\sqrt{ (2(y_o -2h))/(g) }`

we can write the terms of this succession

(n-1) = y₀ -
`(1)/(2)` g t²

t =
`\sqrt{ (2(y_o - (n-1) h))/(g) }`

therefore we see that the times are getting closer as we use each filter, in the expression n would be the number of filters