Question

A body is freely falling from a hight of 30 m.Find the velocity with which it strikes the ground​

Answer 1

Answer:

v = 0

When the body strikes the ground, it is obvious that it will come to rest i e. the final velocity will be zero.

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Answer 2

$\underline{\underline{\pmb{\frak{Answer : }}}}$

24.24 m/s

$\underline{\underline{\pmb{\frak{Explanation : }}}}$

$\textbf{ \textsf{ \large \: Given}} \begin{cases} \sf Height \; (h) = 30 \; m \\ \\ \sf Initial \; Velocity \; (u) = 0 \; m \: s^{-1} \\ \\ \sf Acceleration \; (g) =+ 9.8 \; m \: s^{-2} \\ \end{cases}$

Here, the initial velocity is zero because the ball has been dropped gentry from the height.

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To find : The velocity with which it strikes the ground i.e, final velocity (v).

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$\underline{\bf{ By \; using \; the \;} \rm{ 3rd \; Eq^n} \; \bf{ of \; motion} :}$

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$\bigstar \qquad \underline{\boxed { \pmb{\sf {v}}^{\pmb{\sf {2}}} - \pmb{\sf {u}}^{\pmb{\sf {2}}} = \pmb{\sf {2gh}} }}$

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• v denotes final velocity
• u denotes initial velocity
• g denotes acceleration due to gravity
• h denotes height

$\\ \\ \twoheadrightarrow \qquad \sf { v^2 - \Big (0 \Big )^2 = 2 \times 9.8 \times 30 }$

$\\ \\ \twoheadrightarrow \qquad \sf { v^2 - 0 = 588 }$

$\\ \\ \twoheadrightarrow \qquad \sf { v^2 = 588 }$

$\\ \\ \twoheadrightarrow \qquad \sf { v = \sqrt{588} }$

$\\ \\ \twoheadrightarrow \qquad \underline{\boxed{\pmb{\sf {v = 24.24 }} \; \pmb{\sf{ m\: s^{-1}}} }}$

Therefore, the velocity with which it strikes the ground is 24.24 m/s.

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⠀⠀⠀⠀⠀⠀⠀⠀$\underline{\bf {More \; Information}}$

Equations of motion :

⠀⠀⠀⠀⠀⠀⠀★ v = u + gt

⠀⠀⠀⠀⠀⠀⠀★ h = ut + ½gt²

⠀⠀⠀⠀⠀⠀⠀★ v² – u² = 2gh

Where,

• v denotes final velocity
• u denotes initial velocity
• g denotes acceleration due to gravity
• h denotes height
• t denotes time