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

$\blue{\bold{\underline{\underline{Answer:}}}}$

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

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

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

• According to given question :

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

Answer 2

Answer :

The separation between the ball A and B is 9 m.

Given :

• time = 3s

• Initial speed ( ball A )= 0 m/s

• initial speed ( ball B ) = 3 m/s

To find :

The separation between the thrown balls after 3 s

Formulae Used :

${\bf {\underline{s \: = ut + \frac{1}{2} g {t}^{2} }}}$

• U = initial speed

• t = time

• g = Acceleration due to gravity ( 9.8 = 10 m/s²)

Solution :

Substitute the values in equation ,

For finding the ball (A) :

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

ie, S1 = 45 m

For Ball ( B ) :

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

ie, S2 = 54 m

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

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

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