History, asked by akkhansa, 19 days ago

Given that :–
\sf{ : \implies \: Initial \: Velocity \: (u) = 40}:⟹InitialVelocity(u)=40

\sf{ : \implies \: Final \: velocity \: (v) = 0}:⟹Finalvelocity(v)=0

\sf{: \implies \: Height \: (s) = ? }:⟹Height(s)=?

By applying the third equation of motion :–
\qquad \circ \: { \color{green}{ \underline{ \boxed{ \sf{ {v}^{2} - {u}^{2} = 2gs}}}}}∘
v
2
−u
2
=2gs





we get,
\qquad \leadsto \sf{0 - ( {40}^{2}) = - 2 \times 10 \times s}⇝0−(40
2
)=−2×10×s

\qquad \leadsto \sf{160= - 2 \times 10 \times s}⇝160=−2×10×s

\qquad \leadsto \sf{160= - \: 20 \times s}⇝160=−20×s

\qquad \leadsto \sf{s = \frac{160}{20} }⇝s=
20
160



\qquad \leadsto \sf{s ={ \cancel{ \frac{160}{20} }}}⇝s=
20
160





\qquad \leadsto \sf{s = \: 80 \: \: m/s}⇝s=80m/s

Toatl distance travelled by stone = upward distance + downwars distance
\sf \underbrace{ \mapsto\: 2 \: × \: s \: = \: 2 \: × \: 80 \: = \: 160}
↦2×s=2×80=160



\sf{Total \: Diaplacement =0}TotalDiaplacement =0

\sf{Since \: the \: initial \: and \: final \: point \: is \: same}Sincetheinitialandfinalpointissame



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Answers

Answered by Anonymous
1

Answer:

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Answered by XxkingoftheQueenXx
2

Explanation:

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