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Goryan [66]
3 years ago
7

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

Physics
1 answer:
Zielflug [23.3K]3 years ago
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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Explanation:

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v=\frac{lengthofthewipe}{traveltime} =\frac{1.6}{0.0656} =15.24m/s

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  1. As you now know, water in its natural condition is a colorless, odorless, and tasteless liquid. The hexagonal structure of water's crystals.
  2. The temperature at which a liquid's vapor pressure equals the pressure around it, turning the liquid into vapor, is known as the boiling point. We are aware that water reaches its boiling point at 100°C.
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