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Question

(A) The formula for the horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g}$.Given: $u = 60 \,m \,s^{-1}$, $R = 180 \sqr

Vedclass · Accepted Answer

$30^{\circ}$ or $60^{\circ}$

Vedclass · Answer

(A) The formula for the horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g}$.Given: $u = 60 \,m \,s^{-1}$, $R = 180 \sqrt{3} \,m$, and $g = 10 \,m \,s^{-2}$.Substituting these values into the formula:$180 \sqrt{3} = \frac{(60)^2 \sin(2	heta)}{10}$$180 \sqrt{3} = \frac{3600 \sin(2	heta)}{10}$$180 \sqrt{3} = 360 \sin(2	heta)$$\sin(2	heta) = \frac{180 \sqrt{3}}{360} = \frac{\sqrt{3}}{2}$.Since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$, we have $2	heta = 60^{\circ}$ or $2	heta = 180^{\circ} - 60^{\circ} = 120^{\circ}$.Therefore, $	heta = 30^{\circ}$ or $	heta = 60^{\circ}$.

If the range of a body projected with a velocity of $60 \,m \,s^{-1}$ is $180 \sqrt{3} \,m$, then the angle of projection of the body is (Acceleration due to gravity $= 10 \,m \,s^{-2}$)

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