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prisoha [69]
2 years ago
11

A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate

r at the bottom of the well 5.00 seconds after being dropped from the tower. Calculate the depth of the well. Given: g = -9.81 meters/second2.
Physics
1 answer:
ELEN [110]2 years ago
7 0

A useful formula that gives the free-fall distance from rest in 'T' seconds:

                       D = (1/2 G) x (T²)

      G = 9.81 m/s²
1/2 G = 4.905 m/s²

                       D (5 seconds) = (4.905 m/s²) x (5 sec)²

                                              = (4.905 m/s²) x (25 sec²)

                                              =        122.625 meters .

Since the tower-top is 100m above ground,
the depth of the well, to the top of the water,
accounts for the additional  22.625 meters.

My question is:  How do you know exactly when the stone hit the water ?

You probably stood at the top of the well and listened for the
sound of the 'plop'.  But it took some time after the stone hit
the water for the sound of the plop to come back up to you. 
Well, can't you just subtract that time ?  Yes, but you need
to know how much time to subtract.  That depends on the
depth of the well ... which is exactly what you're trying to
determine, so you don't know it yet.

Oh well.  That's a deep subject.
 
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Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
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a

The orbital speed is v= 2.6*10^{3} m/s

b

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            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

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