A stone is dropped into a deep well and is heard to hit the water 2 s after being dropped. Determine the depth of the well. Assume no time delay in stone hitting the water and sound being heard. Take g= 10 $m s^{- 2}$ .

10 m

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20 m

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30 m

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40 m

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Solution

The correct option is B
20 m

Given, Initial velocity u = 0, $g = 10 m s^{- 2}$ , time t = 2 s

Using second equation of motion, $s = u t + \frac{1}{2} g t^{2}$

$s = 0 + \frac{1}{2} \times 10 \times 2 \times 2$

Hence, depth of well = 20 m.

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A stone is dropped into a deep well and is heard to hit the water 2 s after being dropped. Determine the depth of the well. Assume no time delay in stone hitting the water and sound being heard. Take g= 10 ms−2.

The correct option is B 20 m Given, Initial velocity u = 0, g=10 ms−2, time t = 2 s Using second equation of motion, s=ut+12gt2 s=0+12×10×2×2 Hence, depth of well = 20 m.

A stone is dropped into a deep well and is heard to hit the water 2 s after being dropped. Determine the depth of the well. Assume no time delay in stone hitting the water and sound being heard. Take g= 10 $m s^{- 2}$ .

The correct option is B
20 m

Given, Initial velocity u = 0, $g = 10 m s^{- 2}$ , time t = 2 s

Using second equation of motion, $s = u t + \frac{1}{2} g t^{2}$

$s = 0 + \frac{1}{2} \times 10 \times 2 \times 2$

Hence, depth of well = 20 m.