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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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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 km
Angle of projection, θ = 30°
Acceleration due to gravity, g = 9.8 m/s2
Horizontal range for the projection velocity u0, is given by the relation:
R = u02 Sin 2θ / g
3 = u02 Sin 600 / g
u02 / 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 km
Hence, the bullet will not hit a target 5 km away.

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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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