__Question 4.29__:

A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away ? Assume the muzzle speed to the fixed, and neglect air resistance.

__Solution__:

No

Range,

*R*= 3 km

Angle of projection,

*θ*= 30°

Acceleration due to gravity, g = 9.8 m/s

^{2}

Horizontal range for the projection velocity

*u*

_{0}, is given by the relation:

R = u

_{0}

^{2}Sin 2

*θ /*g

3 = u

_{0}

^{2}Sin 60

^{0}/ g

u

_{0}

^{2}/ g = 2√

*3*.......(i)

The maximum range (

*R*

_{max}) is achieved by the bullet when it is fired at an angle of 45° with the horizontal, that is,

R

_{max}= u

_{0}

^{2}/ g ....(ii)

On comparing equations (

*i*) and (

*ii*), we get:

R

_{max}= 3√

*3*

= 2 X 1.732 = 3.46 km

Hence, the bullet will not hit a target 5 km away.

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