Question

A stone is thrown upwards with an initial velocity of 20 m/s from the ground. Find the time taken by
the stone to reach the ground again. (Take g = 10 SI unit)​

Answer 1

$\displaystyle\large\underline{\sf\red{Given}}$

✭ A stone is thrown upward with an initial velocity of 20 m/s

✭ Gravitational Acceleration = 10 m/s²

$\displaystyle\large\underline{\sf\blue{To \ Find}}$

◈ The time that the ball takes to come back to the ground?

$\displaystyle\large\underline{\sf\gray{Solution}}$

So here we shall find the time that the ball takes to reach its highest point ans then multiply it with 2, as it has to come back again, we shall find the time with the help of the first equation of motion, that is,

$\displaystyle\underline{\boxed{\sf v = u+at}}$

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

➝ $\displaystyle\sf v = u+at$

• v = Final Velocity = 0 m/s
• u = Initial Velocity = 1020 m/s
• a = Acceleration = -10 m/s
• t = Time = ?

➝ $\displaystyle\sf 0 = 20+(-10)\times t$

➝ $\displaystyle\sf -20 = -10t$

➝ $\displaystyle\sf \dfrac{-20}{-10} = t$

➝ $\displaystyle\sf \orange{Time = 2 \ sec}$

So now we shall find the total time taken,

• As the ball goes up [1] & then comes back to the ground [2] so then it would travel twice, so,

➳ $\displaystyle\sf Total \ time = 2\times 2$

➳ $\displaystyle\sf \pink{Total \ time = 4 \ sec}$

$\displaystyle\therefore\:\underline{\sf The \ ball \ will \ come \ back \ in \ 4 \ sec}$

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Answer 2

$\leadsto\mathtt{\bf{\small{\underline {\pink{Required\: answer \: :- }}}}}$

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Given :- }}}}$

• Initial Velocity = 20 m/s
• Gravitational Acceleration = 10 m/s

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ To \: Find :- }}}}$

• Time taken by ball to come back to the ground .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ solution :- }}}}$

$\leadsto$ From First Equation of Motion

$\sf\leadsto \boxed{\sf\purple{ \underset{\blue{\sf \ }}{v} = \underset{\blue{\sf }}{u + at} \ {\blue{\sf }}{{}}}}$

$\leadsto \tt\: according \: to \: the \: question$

$\begin{gathered}\sf \leadsto \: v \: = u \: + \: at \\\\\sf\leadsto 0 = 20 + ( - 10) \times t\\\\\sf\leadsto - 20 = - 10t\\\\\sf\leadsto \frac{ - 20}{ - 10} = t\\\\\sf\leadsto t \: = 2 \: sec\\\\\\\\\\{\pink{\ {} }}\end{gathered}$

To find total time taken as ball goes up and then come back down so then it would travel twice

$\sf\leadsto{{time \: = 2 \times 2 }} \\ \\ \sf\leadsto{{time \: = 4 \: sec }}$

                    ____________