A stone is dropped into a deep well and is heard to hit the water $3.41 s$ after being dropped. Determine the depth of the well.

57.0

meters deep

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57.5

meters deep

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47.0

meters deep

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97.0

meters deep

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Solution

The correct option is A $57.0$ meters deep
Given, acceleration of stone is $a = g = 9.8 m / s^{2}$
Initial velocity is $u = 0 m / s$
Time for hitting water is $t = 3.41 s$
If $d$ be the depth of the well.
Using $d = u t + a t^{2} / 2$ ,
$∴$ $d = 0 (3.41) + 0.5 (9.8) (3.41)^{2} = 56.98 \sim 57 m$

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A stone is dropped into a deep well and is heard to hit the water 3.41s after being dropped. Determine the depth of the well.

The correct option is A 57.0 meters deepGiven, acceleration of stone is a=g=9.8m/s2Initial velocity is u=0m/sTime for hitting water is t=3.41sIf d be the depth of the well. Using d=ut+at2/2,∴ d=0(3.41)+0.5(9.8)(3.41)2=56.98∼57m

A stone is dropped into a deep well and is heard to hit the water $3.41 s$ after being dropped. Determine the depth of the well.

The correct option is A $57.0$ meters deep
Given, acceleration of stone is $a = g = 9.8 m / s^{2}$
Initial velocity is $u = 0 m / s$
Time for hitting water is $t = 3.41 s$
If $d$ be the depth of the well.
Using $d = u t + a t^{2} / 2$ ,
$∴$ $d = 0 (3.41) + 0.5 (9.8) (3.41)^{2} = 56.98 \sim 57 m$