Question

A ball is thrown vertically upward with a velocity of 98 metre per second how long does it take to reach in highest point​

Answer 1

$\sf GivEn \begin{cases} & \sf{Initial\; Velocity,\;u = \bf{98\;m/s}} \\ & \sf{Final\;Velocity,\;v = \bf{0\;m/s}} \end{cases}$

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DIAGRAM:

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$\setlength{\unitlength}{1mm}\begin{picture}(8,2)\thicklines\multiput(9,1.5)(1,0){50}{\line(1,2){2}}\multiput(35,7)(0,4){13}{\line(0,1){2}}\put(10.5,6){\line(3,0){50}}\put(35,60){\circle*{10}}\put(37,7){\large\sf{u = 98 m/s}}\put(37,55){\large\sf{v = 0 m/s}}\put(21,61){\large\textsf{\textbf{Ball}}}\put(43,40){\line(0, - 4){28}}\put(43,35){\vector(0,4){18}} \put(24, - 3){\large\sf{ \sf g = - 9.8 m/s^2 }}\end{picture}$

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

$:\implies\sf v = u + at$

$:\implies\sf 0 = 98 + (- 9.8) \times t$

$:\implies\sf 0 = 98 - 9.8t$

$:\implies\sf 9.8t = 98$

$:\implies\sf t = \cancel{ \dfrac{98}{9.8}}$

$:\implies{\underline{\boxed{\sf{\purple{\;\;10\;s\;\;}}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Time\;taken\;by\;ball\;to\;reach\;highest\;point\;is\; \textbf{10\;s.}}}}$

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

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

$\;\;\;\;\bullet\;\;\sf{v = u + at}$

$\;\;\;\;\bullet\;\;\sf{s = ut + \dfrac{1}{2} at^{2}}$

$\;\;\;\;\bullet\;\;\sf{v^{2} = u^{2} + 2as}$

Answer 2

Answer:

0 = 98 - 9.81(t)

t= 10 s

xf = 980 -490 = 490m This is the maximum height.

Is the maximum height.