Question

A bullet is fired at an angle of 60 deg with vertical with certain velocity hits the ground 3km away. Is it possible to hit the target 5 km away by adjusting the angle of projection?

Answer 1

Given:

A bullet is fired at an angle of 60 deg with vertical with certain velocity hits the ground 3km away.

To find:

Is it possible to hit the target 5 km away by adjusting the angle of projection?

Calculation:

Angle of Projection with horizantal = 90 - 60 = 30°.

$\therefore \: range = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$\implies\: 3000= \dfrac{ {u}^{2} \sin(2 \times {30}^{ \circ} ) }{g}$

$\implies\: 3000= \dfrac{ {u}^{2} \sin( {60}^{ \circ} ) }{10}$

$\implies\:{u}^{2} \sin( {60}^{ \circ} ) = 30000$

$\implies\:{u}^{2} \times \dfrac{ \sqrt{3} }{2} = 30000$

$\implies\:{u}^{2} = \dfrac{60000}{ \sqrt{3} }$

$\implies\:{u}^{2} = 20000 \sqrt{3}$

Now, we know that maximum range of a projectile is obtained at an angle of 45°:

$\therefore \: range \: max= \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$\implies \: range \: max= \dfrac{ {u}^{2} \sin(2 \times {45}^{ \circ} ) }{g}$

$\implies \: range \: max= \dfrac{ {u}^{2} \sin( {90}^{ \circ} ) }{g}$

$\implies \: range \: max= \dfrac{ {u}^{2} }{g}$

$\implies \: range \: max= \dfrac{ 20000 \sqrt{3} }{10}$

$\implies \: range \: max= 2000 \sqrt{3} \: m$

$\implies \: range \: max= 2 \sqrt{3} \: km$

$\implies \: range \: max= 3.46 \: km$

So, the object will not be able to cover distance of 5 km.