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ohaa [14]
2 years ago
15

A stone is thrown vertically downward with an initial speed of 12.0 m/s from the top of a building. The stone takes 1.54 s to re

ach the ground.
a) What is the height of the building?
b) What is the speed of the stone when it hits the ground?
Physics
1 answer:
mr_godi [17]2 years ago
8 0

(a) The stone travels a vertical distance <em>y</em> of

<em>y</em> = (12.0 m/s) <em>t</em> + 1/2 <em>g t</em> ²

where <em>g</em> = 9.80 m/s² is the acceleration due to gravity. Note that this equation assume the downward direction to be positive, and that <em>y</em> = 0 corresponds to the height from which the stone is thrown.

So if it reaches the ground in <em>t</em> = 1.54 s, then the height of the building <em>y</em> is

<em>y</em> = (12.0 m/s) (1.54 s) + 1/2 (9.80 m/s²) (1.54 s)² ≈ 30.1 m

(b) The stone's (downward) velocity <em>v</em> at time <em>t </em>is

<em>v</em> = 12.0 m/s + <em>g t</em>

so that after <em>t</em> = 1.54 s, its velocity is

<em>v</em> = 12.0 m/s + (9.80 m/s²) (1.54 s) ≈ 27.1 m/s

(and of course, speed is the magnitude of velocity)

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An electron is released from rest in a weak electric field given by vector E = -1.30 10-10 N/C ĵ. After the electron has travel
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Answer:

v =  6.45 10⁻³ m / s

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Let's calculate

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              v =√ (0 + 2 13.0 1.6 10⁻⁶)

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4 0
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