Question

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 away? Assume the muzzle speed to be fixed and neglect air resistance.

Answer 1

hi there

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Answer 2

It is possible to hit a target 5.0 Km away by adjusting the angle of projection of the fired bullet.

1. Case 1: when angle of projection is 30 degree.

2. Here range = $u^{2}$*sin (2θ)/g.

3. Now range given is 3000 m. So equating we get:

          i. 3000 = $u^{2}$*sin (2θ)/g.

         ii. 3000*g/sin (2θ) = $u^{2}$

        iii. 3000*9.81/sin(60)  = $u^{2}$

        iv. u= 185 meters per seconds (approx)

4. Now when the range is 5000 m.

      i. 5000=  $u^{2}$*sin (2θ)/g.

     ii. sin(2θ)  = 5000*9.81/185*185

                     =1.433

     iii.     θ= 46 degree (approx)

5. Since this angle is feasible so by adjusting its angle of projection, can one hope to hit a target 5.0 Km away.