Question

a large number of bullets are fired ina ll directions
with smae speed v
what is the maximum area on ground on which these bullets will spread
?
1 πv2/g 2 πv4/G2 3) π2v4/g2

4) π2v2/g2​

Answer 1

A large number of bullets are fired in all directions with same speed v. What is the maximum area on ground on which these bullets will spread?

• $\bf\dfrac{\pi v^2}{g^2}$

• $\bf\dfrac{\pi v^4}{g^2}$

• $\bf\dfrac{\pi^2 v^4}{g^2}$

• $\bf\dfrac{\pi^2 v^2}{g^2}$

Maximum range of bullet will be at :

$\bf \qquad : \implies \: \theta = 45 \degree$

This range will be the radius of the circle whose are should be :

$\qquad : \implies \bf \: \pi {r}^{2}$

$\bold{\qquad : \implies \: r = \dfrac{ {u}^{2} \: sin \: 2 \theta}{g} }$

This is the radius of the circle (maximum) of bullets falling.

Hence, area of the circle will be :

$\qquad : \implies \bf \: a = \pi {r}^{2}$

$\qquad \bf: \implies \: a = \pi \: \bigg({ \dfrac{v ^{2} }{g} } \bigg)^{2}$

$\qquad : \implies \: \bf a = \pi \: \bigg(\dfrac{ {v}^{4} }{ {g}^{2} } \bigg)$

$\qquad : \implies \bf \: a = \dfrac{\pi {v}^{4} }{ {g}^{2} }$

_____________________

Therefore, the correct answer is $\bf\dfrac{\pi v^4}{g^2}$ [0ptionB].

Answer 2

Answer:

A large number of bullets are fired in all directions with same speed v. What is the maximum area on ground on which these bullets will spread?

\bf\dfrac{\pi v^2}{g^2}

g

2

πv

2

\bf\dfrac{\pi v^4}{g^2}

g

2

πv

4

\bf\dfrac{\pi^2 v^4}{g^2}

g

2

π

2

v

4

\bf\dfrac{\pi^2 v^2}{g^2}

g

2

π

2

v

2

\begin{gathered} \\ \end{gathered}

Maximum range of bullet will be at :

\bf \qquad : \implies \: \theta = 45 \degree:⟹θ=45°

This range will be the radius of the circle whose are should be :

\qquad : \implies \bf \: \pi {r}^{2} :⟹πr

2

\bold{\qquad : \implies \: r = \dfrac{ {u}^{2} \: sin \: 2 \theta}{g} }:⟹r=

g

u

2

sin2θ

This is the radius of the circle (maximum) of bullets falling.

Hence, area of the circle will be :

\qquad : \implies \bf \: a = \pi {r}^{2} :⟹a=πr

2

\qquad \bf: \implies \: a = \pi \: \bigg({ \dfrac{v ^{2} }{g} } \bigg)^{2} :⟹a=π(

g

v

2

)

2

\qquad : \implies \: \bf a = \pi \: \bigg(\dfrac{ {v}^{4} }{ {g}^{2} } \bigg):⟹a=π(

g

2

v

4

)

\qquad : \implies \bf \: a = \dfrac{\pi {v}^{4} }{ {g}^{2} } :⟹a=

g

2

πv

4

\begin{gathered} \\ \end{gathered}

_____________________

Therefore, the correct answer is \bf\dfrac{\pi v^4}{g^2}

g

2

πv

4

[0ptionB].