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Answer
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Hint: We need to calculate the time taken by the stone to reach the well at first and deduct that amount from the total time, which is \[3.12s\]. Depth of the well is given, therefore, this remaining time was taken by sound to reach the observer, and hence we can find the
Complete step by step solution:
In the question it is given that the splash is heard after \[3.12s\], this means that the stone reached the well and then the sound travelled back to the observer, this took \[3.12s\].
The following values are given to us:
Time after which the splash is heard \[ = 3.12s\]\[(t)\]
Depth of the well \[ = 45m\]
g \[ = 10m{s^{ - 2}}\]
Let \[{t_1}\] be the time taken by the stone to reach the water.
And \[h\] be the depth of the well \[ = 45m\]
We know, the three equations of motion, from there let us consider the equation which helps us to calculate the distance travelled from initial velocity, acceleration and time taken.
\[h = ut + \dfrac{1}{2}g{t_1}^2\]
Since the stone is thrown, we consider the initial velocity to be zero.
\[\therefore U = 0\]
So, \[h = \dfrac{1}{2}g{t_1}^2\]
Now, rearranging the equation, we get:
\[{t_1} = \sqrt {\dfrac{{2h}}{g}} \]
Putting the values, we obtain:
\[{t_1} = \sqrt {\dfrac{{2 \times 45}}{{10}}} s\]
On solving, we get:
\[{t_1} = 3s\]
Total time taken by the splash to reach the observer is \[ = 3.12s\], therefore, time taken for the sound to travel to the observer is: \[{t_2} = t - {t_1}\]
Therefore, \[{t_2} = (3.12 - 3)s = 0.12s\]
We know. Distance travelled by the sound is depth of the well, which is \[ = 45m\]
Thus, we can calculate the speed of sound, as we know:
$\Rightarrow$ \[Speed = \dfrac{{{\text{Distance travelled}}}}{{{\text{Time Taken}}}}\]
Putting the values, we obtain:
$\Rightarrow$ \[Speed = \dfrac{{45}}{{0.12}}\]
Thus, we finally arrive at:
$\Rightarrow$ \[Speed = 375m/s\]
This is the required solution.
Therefore, option (B) is correct.
Note: SI unit of speed is \[m/s\], so if we have distances given in kilometre and time given I hours, we must convert them to meters and seconds respectively, and not write \[km/hr\], as the unit of speed. Initial velocity of the stone is taken to be zero, as the velocity remains the same, with respect to the reference frame.
Complete step by step solution:
In the question it is given that the splash is heard after \[3.12s\], this means that the stone reached the well and then the sound travelled back to the observer, this took \[3.12s\].
The following values are given to us:
Time after which the splash is heard \[ = 3.12s\]\[(t)\]
Depth of the well \[ = 45m\]
g \[ = 10m{s^{ - 2}}\]
Let \[{t_1}\] be the time taken by the stone to reach the water.
And \[h\] be the depth of the well \[ = 45m\]
We know, the three equations of motion, from there let us consider the equation which helps us to calculate the distance travelled from initial velocity, acceleration and time taken.
\[h = ut + \dfrac{1}{2}g{t_1}^2\]
Since the stone is thrown, we consider the initial velocity to be zero.
\[\therefore U = 0\]
So, \[h = \dfrac{1}{2}g{t_1}^2\]
Now, rearranging the equation, we get:
\[{t_1} = \sqrt {\dfrac{{2h}}{g}} \]
Putting the values, we obtain:
\[{t_1} = \sqrt {\dfrac{{2 \times 45}}{{10}}} s\]
On solving, we get:
\[{t_1} = 3s\]
Total time taken by the splash to reach the observer is \[ = 3.12s\], therefore, time taken for the sound to travel to the observer is: \[{t_2} = t - {t_1}\]
Therefore, \[{t_2} = (3.12 - 3)s = 0.12s\]
We know. Distance travelled by the sound is depth of the well, which is \[ = 45m\]
Thus, we can calculate the speed of sound, as we know:
$\Rightarrow$ \[Speed = \dfrac{{{\text{Distance travelled}}}}{{{\text{Time Taken}}}}\]
Putting the values, we obtain:
$\Rightarrow$ \[Speed = \dfrac{{45}}{{0.12}}\]
Thus, we finally arrive at:
$\Rightarrow$ \[Speed = 375m/s\]
This is the required solution.
Therefore, option (B) is correct.
Note: SI unit of speed is \[m/s\], so if we have distances given in kilometre and time given I hours, we must convert them to meters and seconds respectively, and not write \[km/hr\], as the unit of speed. Initial velocity of the stone is taken to be zero, as the velocity remains the same, with respect to the reference frame.
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