Question

A bullet fired at an angle of 30° with the horizontal hits the ground 4.0 km away. By
adjusting its angle of projection, can one hope to hit a target 5.0 km away ? Assume
the muzzle speed to be fixed, and neglect air resistance.
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Answer 1

Answer :

Bullet can not hit a target 5 km away.

Step-by-step explanation :

we know,

       $\boxed{\bf Range=\frac{u^2sin2\theta}{g} }$

Let the speed of the bullet be "u"

• A bullet fired at an angle of 30° with the horizontal hits the ground 4.0 km away.

                Range (R) = 4 km

            angle of projection (θ) = 30°

By substituting it's values in the above formula,

                  $4=\frac{u^2sin2(30^{\circ})}{g} \\\\ 4\frac{u^2sin60^{\circ}}{g} \\\\ \frac{u^2(\frac{\sqrt{3}}{2})}{g} =4 \\\\ \frac{u^2}{g} =\frac{8}{\sqrt{3}} \\\\ \frac{u^2}{g} =\frac{8}{\sqrt{3}} \times \frac{\sqrt{3} }{\sqrt{3} } \\\\ \frac{u^2}{g} =\frac{8\sqrt{3}}{3}$

• Range is maximum at θ = 45°

           $R_{max}=\frac{u^2sin2(45^{\circ})}{g} \\\\ R_{max} =\frac{u^2sin90^{\circ}}{g} \\\\ R_{max}=\frac{u^2}{g}$

substitute the value of u²/g here,

          Rₘₐₓ    =  $\frac{8\sqrt{3}}{3}$

                     $\simeq\frac{13.856}{3} \\\\ \simeq\text{4.618 km}$

Hence, bullet can not hit the target 5 km away.

Since, the maximum range with the same speed is 4.618 km which is less than 5 km

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