kramtus51

2022-01-12

A brick is dropped from the roof of a tall building. After it has been falling for a few seconds, it falls 40.0 m in a 1.00-s time interval. What distance will it fall during the next 1.00 s? Ignore air resistance.

amarantha41

Beginner2022-01-13Added 38 answers

Take the positive y direction to be downward, so $y-{y}_{0}=40\text{}m$ .For that one second we can imagine the brick to start falling withinitial velocity $v}_{0$ that makes it fall a distance 40 m in just one second . Having known the initial velocity, we can calculate the distance the brick will move in 2 seconds and then we can know the distance moved in the second 1 s interval by subtracting the first 40 m fom the total distance moved in 2 seconds.

$y-{y}_{0}={v}_{0}t+\frac{1}{2}{>}^{2}$

$40={v}_{0}+4.9$

$v}_{0}=35.1\frac{m}{s$

For t=2s

$y-{y}_{0}=35.1\times 2+4.9\times {2}^{2}=89.8$ m

So, the distance moved in the second 1 s interval is:

$d=89.8-40=49.8$

For t=2s

So, the distance moved in the second 1 s interval is:

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