Question

A ball is thrown upward with a velocity of 10 metre per second find the maximum height reached by the ball and the time taken to come back to the thrower (g=10m/s^2

Answer 1

very easy. apply v2-u2=2as.

next time use s=ut+1/2at2

Answer 2

Aɴꜱᴡᴇʀ

$\huge \tt \pink{maximum \: h = 50m} \\ \huge\tt\pink{time = 2s}$

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Gɪᴠᴇɴ

Initial velocity (u) = 10 m/s

Final Velocity (v) = 0 m/s

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ᴛᴏ ꜰɪɴᴅ

Maximum height

Time taken by ball to come back to thrower

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Sᴛᴇᴘꜱ

Use 3rd equation of motion :

$\begin{lgathered}\dashrightarrow \large \boxed{\tt{v^2 \: - \: u^2 \: = \: 2as}} \\ \\ \dashrightarrow {\tt{0^2 \: - \: 10^2 \: = \: 2 \: \times \: -10 \: \times \: s}} \\ \\ \tt{-100 \: = \: -20s} \\ \\ \dashrightarrow \tt{100 \: = \: 20s} \\ \\ \dashrightarrow \tt{s \: = \: \dfrac{100}{20}} \\ \\ \dashrightarrow \tt{s \: = \: 5} \\ \\ \fbox{ \pink{\boxed{\sf{Maximum \: Height \: is \: 5 \: m}}}}\end{lgathered}$

Use 1st equation of motion :

$\begin{lgathered}\dashrightarrow \large {\boxed{\tt{v \: = \: u \: + \: at}}} \\ \\ \dashrightarrow \tt{0 \: = \: 10 \: + \: (-10)t} \\ \\ \dashrightarrow \tt{-10 \: = \: -10t} \\ \\ \dashrightarrow \tt{t \: = \: \dfrac{10}{10}} \\ \\ \dashrightarrow \tt{t \: = \: 1 \: s(to \: reach \: maximum \: height)} \\ \\ \tt{}so \: the \: time \: to \: reach \: back = 1 \times 2 = \huge{}2 \: s\end{lgathered}$

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$\huge{\mathfrak{\purple{hope\; it \;helps}}}$