Question

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 1

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

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

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Answer 2

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

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

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