Question

A particle thrown vectically up with an intial velocity 9m/s from the surface of earth (take=10m/s^2) . The time (in sec.)taken by the particle to reach a height of 4m from the surface second tiime (in second) is

Answer 1

Answer:

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Explanation:

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Answer 2

$\sf\Large{\underbrace{\purple{Answer\implies 1\:sec}}}$

$\sf\large{\underline{\underline{Given:}}}$

$\sf{Innitial\:velocity\:(u) = 9\:m/s}$

$\sf{Height = 4\:m}$

$\sf{Acceleration\:(g) = 10\:m/{s}^{2}}$

$\sf\large{\underline{\underline{To\:Find:}}}$

$\sf{Time\:took\:reach\:at\:the\:height\:of\:4\:m = ....?}$

$\sf\large{\underline{\underline{SoLUtION:}}}$

$\sf{Time\:taken\:by\:the\:particle\:to\:reach\:a\:height\:(h)}$$\sf{from\:the\:surface\:of\:Earth\:is\:obtained\:from..}$

$\sf\red{\implies h = ut - \dfrac{1}{2}g{t}^{2}}$

$\sf{\underline{\underline{Given:}}}$

$\sf{ u = 9\:m/s}$

$\sf{ h = 4\:m}$

$\sf{ g = 10\:m/{s}^{2}}$

$\sf{\underline{\underline{Therefore:}}}$

$\sf{\purple{\implies 4 = 9t - \dfrac{1}{2} \times 10 \times {t}^{2}}}$

$\sf{\orange{\implies 4 = 9t - 5{t}^{2}}}$

$\sf{\green{\implies 5{t}^{2} - 9t + 4 = 0}}$

$\sf{\orange{\implies 5{t}^{2} - 5t - 4t + 4 = 0}}$

$\sf{\purple{\implies 5t(t - 1) - 4(t - 1) = 0}}$

$\sf{\underline{\underline{Therefore:}}}$

$\sf\red{ t = 1\:second\:or\: t = \dfrac{4}{5}\:second}$

$\sf\large{\underline{\underline{Hence:}}}$

$\sf{The\:time\:taken \:by\:the\:particle\:to\:reach\:a\:height\:of}$$\sf{4\:meter\:from\:the\:surface\:second\:time\:is\:1\:second.}$