Question

When a body is dropped from the top of a building, its

velocity was 20 m/s on reaching the ground, find the

height of the building.​

Answer 1

GivEn:

• Velocity of a body on reaching ground = 20 m/s

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To find:

• Height of building

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SoluTion:

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${\underline{\bf{\bigstar\;As\;per\;given\; Question\;:}}}$

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When a body is dropped from the top of a building, its velocity was 20 m/s on reaching the ground.

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$\bf Here \begin{cases} & \text{Initial Velocity, u = 0 m/s } \\ & \text{Final Velocity, v = 20 m/s} \\ & \text{Acceleration, a = 9.8 m/s^2 }\end{cases}$

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Now,

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${\underline{\sf{\bigstar\;Using\;3rd\;equation\;of\;motion\;:}}}$

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$\star\;{\boxed{\sf{\purple{v^2 - u^2 = 2as}}}}$

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$:\implies\sf 20^2 - 0^2 = 2 \times 9.8 \times s$

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$:\implies\sf 400 - 0 = 19.6s$

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$:\implies\sf s = \dfrac{400}{19.6}$

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$:\implies{\underline{\boxed{\bf{\pink{s = 20.41\;m}}}}}\;\bigstar$

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$\therefore$ Hence, The height of tower will be 20.41 m.

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Additional Information:-

★ There are three equations of motion:-

⠀⠀⠀✩ $\sf v = u + at$

⠀⠀⠀✩ $\sf s = ut + \dfrac{1}{2} at^2$

⠀⠀⠀✩ $\sf v^2 - u^2 = 2as$

⠀⠀ ━━━━━━━━━━━━━━━━━━━━━

$\;\;\star\;\sf Acceleration (a) = \dfrac{Final\; velocity (v) - Initial\; velocity (u)}{Time (t)}$

Answer 2

Answer:

✡ Question ✡

➡ When a body is dropped from the top of a building, its velocity was 20 m/s on reaching the ground, find the height of the building.

✡ Given ✡

➡ Velocity of a body on reaching ground = 20 m/s.

✡ To Find ✡

➡ The height of the building.

✡ Formula Used ✡

By using the third equation of motion:-

⭐ v² - u² = 2as ⭐

✡ Solution ✡

Given:-

⚫ Initial Velocity, (u) = 0 m/s

⚫Velocity, (v) = 20 m/s

⚫ Acceleration, (a) = 9.8 m/s²

▶ According to the question :-

Using 3rd law of motion :-

✴ v² - u² = 2as ✴

=> 20² − 0² = 2×9.8×s

=> 400 − 0 = 19.6s

=> s = $\dfrac{400}{19.6}$

=> $\bold{\large{\fbox{\color{green} {s = 20.41 m}}}}$

Explanation:

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