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(C) The formula for the horizontal range $R$ of a projectile is given by $R = \frac{v^2 \sin(2	heta)}{g}$,where $v$ is the initial velocity,$	heta$

Vedclass · Accepted Answer

$100$

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(C) The formula for the horizontal range $R$ of a projectile is given by $R = \frac{v^2 \sin(2	heta)}{g}$,where $v$ is the initial velocity,$	heta$ is the angle of projection,and $g$ is the acceleration due to gravity.Since the velocity $v$ and gravity $g$ are constant,the range is directly proportional to $\sin(2	heta)$,i.e.,$R \propto \sin(2	heta)$.Given $	heta_1 = 15^{\circ}$ and $R_1 = 50\,m$. For $	heta_2 = 45^{\circ}$,we need to find $R_2$.Using the ratio: $\frac{R_1}{R_2} = \frac{\sin(2	heta_1)}{\sin(2	heta_2)} = \frac{\sin(2 	imes 15^{\circ})}{\sin(2 	imes 45^{\circ})} = \frac{\sin(30^{\circ})}{\sin(90^{\circ})}$.Substituting the values: $\frac{50}{R_2} = \frac{0.5}{1} = \frac{1}{2}$.Therefore,$R_2 = 50 	imes 2 = 100\,m$.

The range of the projectile projected at an angle of $15^{\circ}$ with the horizontal is $50\,m$. If the projectile is projected with the same velocity at an angle of $45^{\circ}$ with the horizontal,then its range will be $........\,m$.

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