Physics, asked by snow5050, 5 months ago

Find separation between A and B after 3 sec is 2 balls with initial speed 0 and 3m/s respectively are thrown from a building.

Answer with explantion
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Answers

Answered by Anonymous
3

Solution :

Substitute the values in equation ,

For finding the ball (A) :

\begin{gathered}\sf\implies \: s = u_{1} t+ \frac{1}{2} \: g {t}^{2} \\ \\ \implies\sf \: 0 \times 3 + \frac{1}{2} \times 10 \times {3}^{2} \\ \\ \sf\implies \: 0 + \frac{1}{2} \times 10 \times 9 \\ \\ \sf\implies \: 5 \times 9 \\ \\ \bf \implies 45\end{gathered}

ie, S1 = 45 m

For Ball ( B ) :

\begin{gathered}\sf\implies \: s = u_{2} \: t \: + \frac{1}{2} \: g \: {t}^{2} \\ \\ \sf\implies \: 3 \times 3 + \frac{1}{2} \times 10 \times {3}^{2} \\ \\ \sf\implies \: 9 + \frac{1}{2} \times 10 \times 9 \\ \\ \sf\implies \: 9 + 5 \times 9 \\ \\ \bf\implies \: 54\end{gathered}

ie, S2 = 54 m

Then, To find the separation between them = S2 - S1

\begin{gathered}\bullet \: \: \sf s_{1} \: = 45 \\ \sf\bullet \: \: s_{2} \: = 54\\ \\ \sf \: s_{2} - s_{1} \: = 54 - 45 \\ \sf \: = 9 \: m\end{gathered}

_______________________

Answered by Anonymous
1

\huge\underline{\overline{\mid{\bold{\blue{Answer ♡}}\mid}}}

\green{\tt{\therefore{s_{2} - s_{1} =9 \: m}}}

\pink{\bold{\underline{\underline{Step-by-step\:explanation:}}}}

\begin{gathered}\green{\underline{\bold{Given:}}} \\ \tt: \implies Initial \: speed \: of \: ball \: A = 0 \: m/s \\ \\ \tt: \implies Initial \: speed \: of \: ball \: B= 3 \: m/s \\ \\ \tt : \implies Time = 3 \: sec \\ \\ \red{\underline{\bold{To \: Find:}}} \\ \tt: \implies </p><p>Sepration \: after \: 3 \: sec =?\end{gathered}

Given:

• According to given question :

\begin{gathered}\green{ \star} \tt \:Acceleration = g = 10 \: {m/s}^{2} \\ \\ \bold{As \: we \: know \: that(For \: ball \: A)} \\ \tt: \implies s_{1} = u_{1}t + \frac{1}{2} g {t}^{2} \\ \\ \tt: \implies s_{1} = 0 \times 3 + \frac{1}{2} \times 10 \times {3}^{2} \\ \\ \tt: \implies s_{1} =5 \times 9 \\ \\ \tt: \implies s_{1} =45 \: m - - - - - (1) \\ \\ \bold{Similarly \:( For \:ball \: B)} \\ \tt: \implies s_{2} = u_{2}t + \frac{1}{2} {gt}^{2} \\ \\ \tt: \implies s_{2} = 3 \times 3 + \frac{1}{2} \times 10 \times {3}^{2} \\ \\ \tt: \implies s_{2} = 9 + 5 \times 9 \\ \\ \tt: \implies s_{2} = 9 + 45 \\\\ \tt: \implies s_{2} = 54 \: m - - - - - (2) \\ \\ \text{subtracting \: (1) \: from \: (2)} \\ \\ \tt: \implies s_{2} - s_{1} = 54- 45 \\ \\ \green{\tt: \implies s_{2} - s_{1} =9 \: m}\end{gathered}

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