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.
$T a k e a = 10 \frac{m}{s^{2}}$ .

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, u = 0,
$a = 10 \frac{m}{s^{2}}$ ,
t = 2 s

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

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

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 a=10ms2.

The correct option is B 20 m Given, u = 0, a=10ms2, t = 2 s Using second equation of motion, s=ut+12at2 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.
$T a k e a = 10 \frac{m}{s^{2}}$ .

The correct option is B
20 m

Given, u = 0,
$a = 10 \frac{m}{s^{2}}$ ,
t = 2 s

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

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

Hence, depth of well = 20 m.