Question

If a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the water?

Answer 1

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Given:

Initial velocity (u) = 0m/s

time taken = 3.0 seconds

Acceleration ( here acceleration due to gravity)

= 10 m/s^2 or 9.8m/s^2

so we have to find the distance it travelled.

Here are going to use 2nd equation of motion

$s = ut + \frac{1}{2} a {t}^{2}$

$s = (0)(3) + \frac{1}{2} \times 10 \times 3 \times 3 \\ s = 0 + 5 \times 9 \\ s = 0 + 45 \\ s = 45meters$

So the distance travelled by the stone is 45 meters.

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Answer 2

Answer:

\color{Red}{\huge{\fbox{\fbox{\fbox{\color{Blue}{..!!!Hello\:Mate!!!..}}}}}}

..!!!HelloMate!!!..

Given:

Initial velocity (u) = 0m/s

time taken = 3.0 seconds

Acceleration ( here acceleration due to gravity)

= 10 m/s^2 or 9.8m/s^2

so we have to find the distance it travelled.

Here are going to use 2nd equation of motion

s = ut + \frac{1}{2} a {t}^{2}s=ut+

2

1

at

2

\begin{gathered}s = (0)(3) + \frac{1}{2} \times 10 \times 3 \times 3 \\ s = 0 + 5 \times 9 \\ s = 0 + 45 \\ s = 45meters\end{gathered}

s=(0)(3)+

2

1

×10×3×3

s=0+5×9

s=0+45

s=45meters

So the distance travelled by the stone is 45 meters.

\color{Red}{\underline{\fbox{\fbox{\fbox{\color{Blue}{Mark\:Me\:as\:brainliest}}}}}}

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