Question

31. A ball is thrown vertically upwards with a velocity of 56m/s. Calculate
ball is thrown vertically way
i) The maximum height at which it rises,
in) The total time it takes to return to the surface of the earth.
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Answer 1

$\dag\:\underline{\sf AnsWer :}$

• Initial Velocity (u) = 56 m/s
• Final Velocity (v) = 0 m/s
• Acceleration due to gravity = -10 m/s²

We are asked to find the maximum height to which it rises and the total time it takes to return to the surface of the earth.

$\bigstar \: \underline{\textbf{ Maximum height to which it rises : } }$

• Using third kinematical equation of motion we can find the maximum height to which ball rises :

$:\implies\sf v^2 - u^2 = 2gh$

$:\implies\sf h = \dfrac{v^2- u^2}{2g}$

$:\implies\sf h = \dfrac{(0)^2- (56)^2}{2 \times ( - 10)}$

$:\implies\sf h = \dfrac{- (56)^2}{ - 20}$

$:\implies\sf h = \dfrac{(56)^2}{ 20}$

$:\implies\sf h = \dfrac{3136}{ 20}$

$:\implies \underline{ \boxed{\frak{ h = 156.8 \: m}}}$

$\therefore\:\underline{\textsf{The maximum height to which it rises is \textbf{156.8 m}}}.$

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$\bigstar \: \underline{\textbf{ Time taken : } }$

$\dashrightarrow\:\:\sf v = u + gt$

$\dashrightarrow\:\:\sf v - u = gt$

$\dashrightarrow\:\:\sf \dfrac{v - u }{g}= t$

$\dashrightarrow\:\:\sf t = \dfrac{v - u }{g}$

$\dashrightarrow\:\:\sf t = \dfrac{0 - 56}{ - 10}$

$\dashrightarrow\:\:\sf t = \dfrac{ - 56 }{ - 10}$

$\dashrightarrow\:\:\sf t = \dfrac{56 }{10}$

$\dashrightarrow\:\: \underline{ \boxed{\frak{ t = 5.6 \: s}}}$

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$\bigstar \: \underline{\textbf{ The total time it takes to return to the surface of the earth : } }$

$\longrightarrow\:\:\textsf {Total time = Time of ascesnd + Time of descend}$

$\longrightarrow\:\:\textsf {Total time = 5.6 + 5.6}$

$\longrightarrow\:\: \underline{ \boxed{\frak {Total \: time = 11.2 \: s}}}$

$\therefore\:\underline{\textsf{The total time it takes to return to the surface of the earth is \textbf{ 11.2 second}}}.$

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Answer 2

DIAGRAM:

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$\setlength{\unitlength}{1mm}\begin{picture}(8,2)\thicklines\multiput(9,1.5)(1,0){50}{\line(1,2){2}}\multiput(35,7)(0,4){13}{\line(0,1){2}}\put(10.5,6){\line(3,0){50}}\put(35,60){\circle*{10}}\put(37,7){\large\sf{u = 56 m/s}}\put(37,55){\large\sf{v = 0 m/s}}\put(21,61){\large\textsf{\textbf{Ball}}}\put(43,40){\line(0, - 4){28}}\put(43,35){\vector(0,4){18}} \put(24, - 3){\large\sf{ \sf g = - 9.8 m/s^2 }}\put(48,30){\large\sf{H = ?}}\end{picture}$

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$\frak{Given} \begin{cases} \sf Initial\:velocity\:of\:ball,\:(u) = \frak{56\:m/s} & \\ \\ \sf Final\:velocity\:of\:ball,\:(v) = \frak{0\:m/s}&\end{cases}$

${\underline{\frak{Need\:to\:find\::}}}$

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• i) The maximum height at which it rises,
• ii) The total time it takes to return to the surface of the earth.

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❍ Let's consider the time taken is "t" to reach the maximum height "H".

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Now, By using the 3rd eqⁿ of motion,

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$\star\:{\underline{\boxed{\frak{\purple{v^2 - u^2 = 2gH}}}}}\\\\\\ \bf{\dag}\:{\underline{\frak{Putting\:given\:values\:in\:formula,}}}\\\\\\ :\implies\sf (0)^2 - (56)^2 = 2 \times (-9.8) \times H\\\\\\ :\implies\sf 0 - 3136 = - 19.6 \times H\\\\\\ :\implies\sf - 3136 = - 19.6 H\\\\\\ :\implies\sf H = \cancel{\dfrac{- 3136}{ - 19.6}}\\\\\\ :\implies{\underline{\boxed{\frak{H = 160\:m}}}}\:\bigstar$

$\therefore\:{\underline{\sf{The\:maximum\:height\:at\:which\:the\:ball\:rises\:is\: {\textsf{\textbf{160\:m}}}.}}}$

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Again, By using 1st eqⁿ of motion,

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$\star\:{\underline{\boxed{\frak{\purple{v = u + g \times t}}}}}\\\\\\ \bf{\dag}\:{\underline{\frak{By\:Putting\:given\:values\:in\:formula,}}}\\\\\\ :\implies\sf 0 = 56 + (-9.8) \times t\\\\\\ :\implies\sf 0 = 56 - 9.8t\\\\\\ :\implies\sf 9.8t = 56\\\\\\ :\implies\sf t = \cancel{\dfrac{56}{9.8}}\\\\\\ :\implies{\underline{\boxed{\frak{t = 6\:sec\:(approx.)}}}}\:\bigstar$

$\therefore\:{\underline{\sf{The\:total\:taken\:to\:reach\:the\:ground\:is\: (6 + 6) = {\textsf{\textbf{12\:sec}}}.}}}$