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sveta [45]
2 years ago
10

You and your friends find a rope that hangs down 19m from a high tree branch right at the edge of a river. You find that you can

run, grab the rope, swing out over the river, and drop into the water. You run at 2.0 m/s and grab the rope, launching yourself out over the river.How long must you hang on if you want to drop into the water at the greatest possible distance from the edge?
Physics
1 answer:
tangare [24]2 years ago
5 0

Answer:

Explanation:

Given

length of rope L=19\ m

velocity while running v=2\ m/s

when the person jumps off the bank and hang on the rope then we can treat the person as pendulum with Time period T which is given by

T=2\pi \sqrt{\frac{L}{g}}

T=2\pi \sqrt{\frac{19}{9.8}}

T=2\pi \times 1.392

T=8.74\ m/s

Greatest Possible distance will be covered when person reaches the other extreme end  of assumed pendulum (velocity=zero)

therefore he must hang on for 0.5 T time

time=0.5\times 8.74=4.37\ s

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A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp
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Answer:

the required value is x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

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substitute the value of m =1kg, k = 21N/M, and \beta = 10

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suppose the equation of the form x =e^m^t,

and the auxilliary equation is given by

m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3

The general solution for the above differential equation is

x(t) =C_1e^{-3t}+C_2e^{-7t}

Derivate with respect to t

x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

x'(0) = 0

substitute the initial condition

C_1 +C_2 = 1

-3C_1-7C_2=0

Solve the above equation to get C₁ and C₂

C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}

substitute for C₁ and C₂ in general solution

x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

Thus the required value is x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

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