Question

A bullet fired at an angle 30^(@) with the horizontal hits the ground 5.0km away.To decrease the range of bullet to 3km keeping muzzle speed constant the angle of projection should be​

Answer 1

Let u be the initial velocity of projection and g be the acceleration due to gravity.

Range is generally defined mathematically as,

$\star \: \boxed{ \boxed{ \sf r = \dfrac{u {}^{2} \sin(2 \alpha ) }{g} }}$

When range is 5Km and angle of projection is 30 degrees.

$\sf 5= \dfrac{u {}^{2} \sin(60) }{g} - - - - (1)$

When range is 3Km,

$\sf 3= \dfrac{u {}^{2} \sin(2 \gamma ) }{g} - - - - (2)$

Dividing both the equations,

$\longrightarrow \sf \: \dfrac{5}{3} = \dfrac{ \sin(60) }{ \sin(2 \gamma ) } \\ \\ \longrightarrow \sf \dfrac{5}{3} = \dfrac{ \sqrt{3} }{2 \sin( 2\gamma ) } \\ \\ \longrightarrow \sf \dfrac{10}{3 \sqrt{3} } = \dfrac{1}{ \sin( 2 \gamma ) } \\ \\ \longrightarrow \sf \sin(2 \gamma ) = \dfrac{3 \sqrt{3} }{10} \\ \\ \longrightarrow \sf 2 \gamma = sin {}^{ - 1} \dfrac{3 \sqrt{3} }{10} \\ \\ \longrightarrow \underline{ \boxed{ \sf \gamma = \dfrac{1}{2} {sin}^{ - 1} \dfrac{3 \sqrt{3} }{10} }}$

Option (C) is correct..