A bullet fired at an angle of 30° with the horizontal hits the ground 3 km away. By adjusting its angle of projection can one hope to hit a target 5 km away. Assume the nuzzle speed to be fixed and neglect air resistance.

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Rajan Kumar mishra

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Range, R = 3 kmAngle of projection, θ = 30°Acceleration due to gravity, g = 9.8 m/s2Horizontal range for the projection velocity u0, is given by the relation:R = u02 Sin 2θ / g3 = u02 Sin 600 / gu02 / g = 2√3 …….(i)The maximum range (Rmax) is achieved by the bullet...
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Range,R= 3 kmAngle of projection,θ= 30°Acceleration due to gravity, g = 9.8 m/s2Horizontal range for the projection velocityu0, is given by the relation:R = u02Sin 2θ /g3 = u02Sin 600/ gu02/ g = 2√3 …….(i)The maximum range (Rmax) is achieved by the bullet when it is fired at an angle of 45° with the horizontal, that is,Rmax= u02/ g ….(ii) On comparing equations (i) and (ii), we get: Rmax= 3√3= 2 X 1.732 = 3.46 kmHence, the bullet will not hit a target 5 km away. read less
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R=U²sin2@/g U=√(R×g/sin2@) U=√(3000×9.8/sin60°) U=184.25 For maximum range@=45° R (max)=U/g=3464 So it is not possible to reach the bullet on given target
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