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(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2	heta)}{g}$.Given: Initial velocity $u = 10 \,m \,s^{-1}$,

Vedclass · Accepted Answer

$314$

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(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2	heta)}{g}$.Given: Initial velocity $u = 10 \,m \,s^{-1}$,angle of projection $	heta = 45^{\circ}$,and acceleration due to gravity $g = 10 \,m \,s^{-2}$.Substituting the values: $R = \frac{10^2 \sin(2 	imes 45^{\circ})}{10} = \frac{100 	imes \sin(90^{\circ})}{10} = \frac{100 	imes 1}{10} = 10 \,m$.Since the bullets are fired in all possible directions,they will land on a circle of radius $R = 10 \,m$ on the ground.The area $A$ covered by the bullets is the area of this circle: $A = \pi R^2$.$A = \pi 	imes (10)^2 = 100\pi$.Using $\pi \approx 3.14$,we get $A = 100 	imes 3.14 = 314 \,m^2$.

If bullets are fired in all possible directions from the same point with an equal velocity of $10 \,m \,s^{-1}$ and an angle of projection $45^{\circ}$,then the area covered by the bullets on the ground is nearly (Acceleration due to gravity $g = 10 \,m \,s^{-2}$) (in $\,m^2$)

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