Question

# A bullet is fired from horizontal ground at some angle passes through the point ( left(frac{3 mathrm{R}}{4}, frac{mathrm{R}}{4}right), ) where "R' is the range of the bullet. Assume point of the fire to be origin and the bullet moves in ( x ) -y plane with ( x ) -axis horizontal and ( y ) -axis vertically upwards. Angle of projection is ( frac{alpha pi}{180} ) radian. Find ( alpha: )

Solution

( quad y=x tan theta-frac{1}{2} frac{g x^{2}}{u^{2} cos ^{2} theta} )

( Rightarrow frac{R}{4}=frac{3 R tan theta}{4}-frac{9 R^{2} g}{32 u^{2} cos ^{2} theta} )

( Rightarrow frac{R}{4}=frac{3 R}{4}left(tan theta-frac{3 R g}{left.theta^{2} cos ^{2} thetaright)}right. )

( Rightarrow frac{1}{3}=tan theta-frac{3 g}{theta^{2} cos ^{2} theta} cdot frac{2 u^{2} sin theta cos theta}{g} )

( Rightarrow quad frac{1}{3}=tan theta-frac{3}{4} tan theta Rightarrow frac{1}{3}=frac{tan theta}{4} )

( Rightarrow quad tan theta=frac{4}{3} Rightarrow theta=53^{circ} )

( begin{aligned} Rightarrow theta &=frac{53 pi^{c}}{180}=frac{alpha pi}{180} & Rightarrow quad alpha=53 end{aligned} )