Question

(c) A stone is thrown across the frozen surface of a lake and it comes to rest after traveling a distance of 50 m. Find its retardation.​

Answer 1

Given : A stone of 1 kg is thrown with a velocity of 20m/s  

across the frozen surface of a lake and comes to rest after travelling a distance of 50m.

To Find : Retardation

Solution:

u = 20 m/s

S = 50 m

V = 0   at rest

V² - U² = 2aS

=> 0² - 20² = 2a (50)

=> - 400 = 100a

=>  a = - 4

-ve sign shows retardation

retardation is  4 m/s²

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Answer 2

Explanation:

★ Correct Question :-

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Q.] 1 kg of stone is thrown vertically with a velocity of 20 m/s across the frozen surface of lake and it comes to rest after traveling a distance of 50 m. Find its Retardation.

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★ Given :-

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• lnitial Velocity [ U ] = 20 m/s
• Distance Travelled [ S ] = 50 m
• Final Velocity [ V ] = 0

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★ To Find :-

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• Retardation of the stone.

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★ Answer :-

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⠀⠀⠀⠀⠀$\maltese \: \bold \red{Formula \: Required :- } \\ \\ \large\mapsto{ \bold{ {v}^{2} - {u}^{2} = 2 a\:S}}$

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★ Solution :-

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• Substituting the given values, in the above formula :-

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⠀⠀⠀⠀⠀$\footnotesize \bold { \longrightarrow {0}^{2} - {20}^{2} = 2a \: (50) }\\ \\ \footnotesize{\longrightarrow \bold{ - 400 = 100 \: a}} \\ \\ \footnotesize\longrightarrow\bold {a = \cancel{ \frac{ - 400}{100} } } \\ \\ \boxed { \mapsto \bold \red{ \: - 4}}$

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★ Hence,

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• Retardation of stone = $\bold\pink {4 m/s^2}$

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