Question

When a stone is dropped from a building reaches the ground after 2 s. Find out the height of the building. (acceleration due to gravity g = 10 m / s)​

Answer 1

ans is 20m you can check solution in attachment

Answer 2

✴ Given :

• A stone is dropped from a building reaches the ground in 2s. (Acceleration due to gravity g = 10 m/s ).

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✴ To Find :

• Find the height of the building.

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✴ Solution :

Formula Used :

$\large{\gray{\bigstar}} \: \: \: {\underline{\boxed{\red{\sf{ S = ut + \dfrac{1}{2} gt²}}}}}$

Where :

• ➳ S = Height of building = ?
• ➳ U = Initial velocity = 0 m/s
• ➳ T = Time taken = 2s
• ➳ G = Acceleration due to gravitation = 10 m/s .

$\qquad{━━━━━━━━━━━━━━━━━━━━━━━━━━}$

Finding the Height of building :

${:\implies{\qquad{\sf{ S = ut + \dfrac{1}{2}gt²}}}}$

${:\implies{\qquad{\sf{ S = 0 \times 2 + \dfrac{1}{2} \times 10 \times (2)²}}}}$

${:\implies{\qquad{\sf{ S = 0 \times 2 + \dfrac{1}{\cancel2} \times \cancel{10} \times (2)²}}}}$

${:\implies{\qquad{\sf{ S = 0 \times 2 + 1 \times 5 \times (2)²}}}}$

${:\implies{\qquad{\sf{ S = 0 \times 2 + 1 \times 5 \times 4}}}}$

${:\implies{\qquad{\sf{ S = 0 + 1 \times 20}}}}$

${:\implies{\qquad{\sf{ S = 0 + 20}}}}$

$\large{\qquad \: \: \: {\red{:\implies}}{\underline{\boxed{\color{darkblue}{\sf{ 20 \: m}}}}}}{\red{\bigstar}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━━━━━━━}$

Therefore :

❝ Height of the building is 20 m . ❞

${\orange{\underline{\pink{▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬}}}}$

✴ Know More :

3 Laws of motion :

$\begin{gathered}\red{\large \qquad \boxed{\boxed{\begin{array}{cc} \ ➳ \: \: {\orange{\bf {v = u + at}}} \\ \\ \ ➳ \: \: {\orange{\bf {s = ut + \dfrac{1}{2}g {t}^{2}}}} \\ \\ \ ➳ \: \: {\orange{\bf{{v}^{2} - {u}^{2} = 2as}}}\end{array}}}}\end{gathered}$