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TiliK225 [7]
3 years ago
15

If an egg is thrown up and off the roof of a 100 foot building with an initial upward velocity of 10 feet per second, how long u

ntil it hits the ground?
Mathematics
1 answer:
Deffense [45]3 years ago
8 0
The equation of the height as a function of time for this case is given by:
 h (t) = - 16t ^ 2 + 10t + 100
 By the time the egg hits the ground we have:
 -16t ^ 2 + 10t + 100 = 0
 We look for the roots of the polynomial:
 t1 = -2.2069555463432966
 t2 = 2.8319555463432966
 As it is about time, we use the positive root:
 t = 2.83 s
 Answer:
 
it hits the ground at:
 
t = 2.83 s
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Answer:

correct option is C)  2.8

Step-by-step explanation:

given data

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in water loop formed =  10 loops

solution

we consider  mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = \rho

case (1)  

when 8 loop form with 2 adjacent node is \frac{\lambda }{2}

so here

l = \frac{8 \lambda _1}{2}      ..............1

l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}

and we know velocity is express as

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\sqrt{\frac{Tension}{mass\ per\ unit \length }}   =   f × \lambda_1

here tension = mg

so

\sqrt{\frac{mg}{\mu}}   =   f × \lambda_1     ..........................3

and

case (2)  

when 8 loop form with 2 adjacent node is \frac{\lambda }{2}

l = \frac{10 \lambda _1}{2}      ..............4

l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}

when block is immersed

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T + v × \rho × g = mg

and

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from equation 2

f × \lambda_2 = f  × \frac{1}{5}  

\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}     .......................5

now we divide eq 5 by the eq 3

\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}

solve irt we get

1 - \frac{\rho _{stone}}{\rho _{water}}  = \frac{16}{25}

so

relative density \frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}

relative density = 2.78 ≈ 2.8

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Answer:

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