If a cell phone is dropped from a very tall building, how far has the phone fallen after 2.7 seconds, neglecting air resistance?

1 answer:

6 0

The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
$g=9.81 m/s^2$

So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by
$y(t) = \frac{1}{2}gt^2$
And if we substitute t=2.7 s, we find the distance covered:
$y(t)= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2=35.8 m$

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For a better understanding let's take a look at the attached image.

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