Question

A ball is thrown vertically upward with a velocity of 49 m/s, find the maximum height attained by the ball.​

Answer 1

Given :-

• initial velocity=49 m/s
• final velocity=0 m/s

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To Find maximum height attained by given ball.

$\begin{gathered} \qquad {\rule{200pt}{2pt}} \end{gathered}$

${\Large \: {\underline{\underline{\gray{\sf{Solution:-}}}}}}$

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Formula Using

$\boxed {\begin{array}{ccc} \bf Using\:3^{rd}\:equation\:of\:motion \\ \\ \sf v^2-u^2=2gs \\ \\ \sf We\:know\:that\:g=9.8\:m/s\end{array}}$

Here

• v=final velocity
• u=initial velocity
• g=gravity
• s=distance travelled

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${\large{\bigstar \: {\underline{\sf{Calculation\: :-}}}}}$

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Calculating Height attained by 3rd equation of motion.

$\begin{gathered} {\rule{170pt}{2pt}} \end{gathered}$

${\dashrightarrow{\sf{\qquad v^2-u^2=2gs}}}$

Putting Values...

${\dashrightarrow{\sf{\qquad (0)^2-(49)^2=2\times (-9.8) \times s}}}$

Simplifying Values...

${\dashrightarrow{\sf{\qquad (-49)^2=2\times (-9.8) \times s}}}$

${\dashrightarrow{\sf{\qquad -2401=-19.6 \times s}}}$

Substituting Values...

${\dashrightarrow{\sf{\qquad s=\dfrac{-2401}{-19.6}}}}$

${\dashrightarrow{\sf{\qquad s=\dfrac{\bcancel{-}2401}{\bcancel{-}19.6}}}}$

${\dashrightarrow{\sf{\qquad s=\cancel\dfrac{2401}{19.6}=122.5}}}$

Getting the answer...

${\leadsto\qquad{\underline{\boxed{\pmb{\tt{Height\:attained\:by\:ball\:is\:122.5\:meters}}}}\pink\bigstar}}$

$\begin{gathered} \qquad {\rule{200pt}{2pt}} \end{gathered}$

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${\therefore}$Height attained by the ball is 122.5 meters

$\begin{gathered} {\underline{\rule{200pt}{3pt}}} \end{gathered}$

Additional Information :-

$\boxed {\begin{array}{ccc} \bf \underline{Equations\:of\:motion} \\ \\ \sf \odot 1^{st}\:equation\:of\:motion,\:v=u+at\\ \\ \sf \odot 2^{nd}\:equation\:of\:motion,\:s=\dfrac{1}{2}at^2 \\ \\ \sf \odot 3^{rd}\:equation\:of\: motion,\:v^2+u^2=2as\end{array}}$

Answer 2

Given : The Ball is thrown upward with a 49 m/s .

To Find : Find the distance Covered by Ball

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SolutioN :

$\dag$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ {v}^{2} - {u}^{2} = 2gs }}}}}$

Where :

• v = Final Velocity
• u = Initial Velocity
• g = Force of Gravitation
• s = Distance Covered

$\dag$ Calculating the Distance Covered :

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { {v}^{2} - {u}^{2} = 2gs } \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { {0}^{2} - {49}^{2} = 2 \times \bigg( - 9.8 \bigg) \times s } \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { 0 - 2401 = - 19.6 \times s } \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \cancel{-} 2401 = \cancel{-} 19.6 \times s } \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { 2401 = 19.6 \times s } \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \dfrac{2401}{19.6} = s } \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \dfrac{24010}{196} = s } \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \cancel\dfrac{24010}{196} = s } \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\pmb{\sf { Distance \; Covered = 245 \; m }}}}} \; {\red{\pmb{\bigstar}}} \\ \\ \end{gathered}$

$\therefore \;$ Distance Covered by the Ball is 245 m .

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