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Alina [70]
3 years ago
11

At t = 0, object A is dropped from the roof of a building. At the same instant, object B is dropped from a window 10 m below the

roof. During their descent to the ground what happens to the distance between the two objects?
Physics
1 answer:
dsp733 years ago
4 0

Answer:

Explanation:

Given two objects are dropped simultaneously

Object A is 10 m higher than object B therefore

Distance covered by object A is given by

y_a(t) is given by

y=ut+\frac{1}{2}at^2

where y=displacement

u=initial velocity

a=acceleration

t=time

y_a(t)=0+0.5gt^2--1

for object B

y_b(t)=0+0.5gt^2--2

Subtract 1 and 2 we get

y_a(t)-y_b(t)=0

i.e. they will travel equal distance in equal time and distance between them remain 10 m until object B hits the ground

           

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Answer:

Explanation:

Givens

Vi = 10 m/s

Vf = 40 m/s

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Formula

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Solution

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3*t = 30                     Divide by 3

3t/3 = 30 / 3

Answer: t = 10 seconds.

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