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Question : ​ A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g= 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone? {\orange{\underline{\bf{Note :
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Answer 1

$\huge{ \boxed{ \tt \pink{Given}}} \downarrow$

$\bf{Initial \: Velocity \: u \: \longrightarrow \: 40}$

$\bf{Fianl \: velocity \: v \: \longrightarrow \: 0}$

$\huge{ \boxed{ \tt \pink{Find}}} \downarrow$

$\bf{Height, \:  s \: \longrightarrow \: ?}$

$\huge{ \boxed{ \tt \pink{Solution}}} \downarrow$

$\bf{By \: third \: equation \: of \: motion}$

$\bf{ {v}^{2} \: - \: {u}^{2} \: \leadsto \: 2gs }$

$\bf{0 \: - \: {40}^{2} \: \leadsto \: - 2 \: \times \: 10 \: \times \: s}$

$\bf{s \: \leadsto \: \frac{160}{20} }$

${\bf\mapsto \: s \: \leadsto \: 80m/s}$

Total distance travelled by stone = upward distance + downwars distance = 2 × s = 160m.

Total Diaplacement = 0, Since the initial and final point is same.

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Answer 2

Answer:

तू कोणत्या एक्साम ची तयारी करतेस भावना...