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(B) The formula for the horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g}$.Given values are:Initial velocity $u = 500

Vedclass · Accepted Answer

$12.5 	imes 10^3 \, m$

Vedclass · Answer

(B) The formula for the horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g}$.Given values are:Initial velocity $u = 500 \, m/s$Angle of projection $	heta = 15^\circ$Acceleration due to gravity $g = 10 \, m/s^2$Substituting these values into the formula:$R = \frac{(500)^2 	imes \sin(2 	imes 15^\circ)}{10}$$R = \frac{250000 	imes \sin(30^\circ)}{10}$Since $\sin(30^\circ) = 0.5$,we have:$R = \frac{250000 	imes 0.5}{10} = 25000 	imes 0.5 = 12500 \, m$This can be written as $12.5 	imes 10^3 \, m$.Therefore,the correct option is $B$.

$A$ bullet is fired from a cannon with a velocity of $500 \, m/s$. If the angle of projection is $15^\circ$ and $g = 10 \, m/s^2$,then the horizontal range is:

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