Physics, asked by Anonymous, 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.​

Answers

Answered by Anonymous
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}

_______________________

Answered by Harsh8557
7

Answer:

  • 9 \: m

Explanation:

Given:-

  • Initial speed of ball (A) = 0 m/s

  • Initial speed of ball (B) = 3 m/s

  • Acceleration (g = 10m/s²)

ToFind:-

  • Sepration after 3 seconds

 Solution :-

For \ ball \ (A)

 \implies \:\:s_{1} =   u_{1}t +  \frac{1}{2} g {t}^{2}

 \implies \:\: s_{1} = 0 \times 3 +   \frac{1}{2}  \times 10 \times  {3}^{2}

 \implies \:\:s_{1} =5 \times 9

 \implies \:\: s_{1} =45 \: m. . . . . (1)\\\\

For \ ball \ (B)

 \implies \:\: s_{2}  =  u_{2}t +  \frac{1}{2}  {gt}^{2}

 \implies \:\:  s_{2}  =  3 \times 3 +  \frac{1}{2}  \times 10 \times  {3}^{2}

 \implies \:\: s_{2}  = 9 + 5 \times 9

 \implies \:\: s_{2}  = 9 + 45

\implies \:\: s_{2}  =54\\\\

Subtracting \: (1) \: from \: (2)

 \implies \:\:  s_{2} - s_{1} = 54- 45

 \implies \:\: s_{2} - s_{1} = 9\:m

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