Question

A stone is dropped vertically upward direction with the velocity of 5m. if the acceleration of the stone during it's motion is 10 m /s in the download direction , what will be the height attained by the stone and much time will it take to reach there. ​

Answer 1

Correct Question:

⠀⠀⠀⠀⠀⠀A stone is dropped vertically upward direction with the velocity of 5m/s. if the acceleration of the stone during it's motion is 10 m/s² in the download direction , what will be the height attained by the stone and much time will it take to reach there.

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DIAGRAM:

⠀

$\setlength{\unitlength}{1mm}\begin{picture}(8,2)\thicklines\multiput(9,1.5)(1,0){50}{\line(1,2){2}}\multiput(35,7)(0,4){13}{\line(0,1){2}}\put(10.5,6){\line(3,0){50}}\put(35,60){\circle*{10}}\put(37,7){\large\sf{u = 5 m/s}}\put(37,55){\large\sf{v = 0 m/s}}\put(18,61){\large\textsf{\textbf{Stone}}}\put(30,40){\vector(0, - 4){32}}\put(30,35){\vector(0,4){22}}\put(16,35){\large\sf{h = ? }}\end{picture}$

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$\frak{Here}\begin{cases} & \sf{Initial\; Velocity,\;u = \bf{5\;m/s}} \\ & \sf{Final\; Velocity\;v = \bf{0\;m/s}} \\ & \sf{Acceleration,\;a = \bf{10\;m/s^2}} \end{cases}$

${\underline{\sf{\bigstar\;Using\;3rd\;equation\;of\;motion\;:}}}$

$\star\;{\boxed{\sf{\purple{v^2 - u^2 = 2as}}}}$

$:\implies\sf 0^2 - 5^2 = 2 \times (-10) \times s$

$\qquad \: \: :\implies\sf - 25 = - 20s$

$\qquad \quad \: :\implies\sf s = \cancel{ \dfrac{25}{20}}$

$\quad\quad:\implies{\boxed{\frak{\pink{s = 1.25\;m}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Height\;obtained\;by\;stone\;is\; \bf{1.25\;m}.}}}$

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${\underline{\sf{\bigstar\;Now,\;Using\;1st\;equation\;of\;motion\;:}}}$

$\star\;{\boxed{\sf{\purple{v = u + at}}}}$

$:\implies\sf 0 = 5 + (- 10) \times t$

$\quad:\implies\sf 0 = 5 - 10t$

$\qquad:\implies\sf 10t = 5$

$\qquad:\implies\sf t = \cancel{ \dfrac{5}{10}}$

$\quad:\implies{\boxed{\frak{\pink{t = 0.5\;s}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;Time\; taken\;by\;stone\;to\;reach\;there\;is\; \bf{0.5\;s}.}}}$

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$\qquad\qquad\qquad\boxed{\bf{\mid{\overline{\underline{\bigstar\:More\:to\:know: \: \bigstar}}}}\mid}$

${\underline{\sf{\bigstar\;There\;are\;three\;equations\;of\;motion\;:}}}$

⠀⠀⠀✩ $\sf v = u + at$

⠀⠀⠀✩ $\sf s = ut + \dfrac{1}{2} at^2$

⠀⠀⠀✩ $\sf v^2 - u^2 = 2as$

⠀⠀ ━━━━━━━━━━━━━━━━━━━━━

$\;\;\star\;\sf Acceleration (a) = \dfrac{Final\; velocity (v) - Initial\; velocity (u)}{Time (t)}$

Answer 2

Given:-

• u = 5m/s
• a = -10m/s²
• v = 0m/s

to find:-

• height
• time

formula:-

3rd equation of motion:-

$\huge \fbox \blue{v²-u²=2as}$

solution:-

0²-5²=2(-10)s

-25 = -20s

s = 1.25 m

by 1st equation of motion:-

$\huge \fbox \blue{v = u+at}$

0 = 5 + (-10)t

0 = 5 - 10t

t = 0.5 sec.