### Q.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.

**Answer:-**

Range,

Angle of projection,

Acceleration due to gravity, g = 9.8 m/s

Horizontal range for the projection velocity

R = u

3 = u

u

The maximum range (

R

*R*= 3 kmAngle 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 /*g3 = u

_{0}^{2}Sin 60^{0}/ gu

_{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.