The well will be BOOM

Classical Mechanics Level 2

A stone is dropped into a well in which the water is H below the top of well. If V is the velocity of sound then time T after which the splash is heard is

2 H V \frac{2H}{V} insufficient data 2 H g \sqrt{\frac{2H}{g}} + H V \frac{H}{V} 2 H V \sqrt{\frac{2H}{V}} + H g \frac{H}{g}

Manish Mayank by Manish Mayank , 20, India

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1 solution

Bernardo Sulzbach
Jun 22, 2014

The answer is the sum of the time the stone takes to hit the water (t1) and the time the sound takes to reach your ears (t2).

H = 0.5 ⋅ g ⋅ ( t 1 ) 2 H=0.5\cdot{}g\cdot{}(t_1)^2

2 ⋅ H ⋅ g − 1 = ( t 1 ) 2 2\cdot{}H\cdot{}g^{-1}=(t_1)^2

t 1 = ( 2 ⋅ H ⋅ g − 1 ) t_1=\sqrt{\left(2\cdot{}H\cdot{}g^{-1}\right)}

t 2 = H ⋅ V − 1 t_2=H\cdot{}V^{-1}

T = ( 2 ⋅ H ⋅ g − 1 ) + H ⋅ V − 1 T=\sqrt{\left(2\cdot{}H\cdot{}g^{-1}\right)}+H\cdot{}V^{-1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

But you don't know where the stone is being dropped from...??

Vamsi Saladi - 6 years, 11 months ago

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I agree. More information could be given. But if the question poster expected us to consider that the stone starts 1 m inside the well or 2 m over it he would have stated it.

Bernardo Sulzbach - 6 years, 11 months ago

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