Question

A shell fired at an angle of 15° to the horizontal hits the ground 3 km away. At what angle it should be projected so that it can hit a target 8 km away?

Answer 1

Given:

A shell fired at an angle of 15° to the horizontal hits the ground 3 km away.

To find:

At what angle it should be projected so that it can hit a target 8 km away?

Calculation:

General experience for range of Projectile:

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

In the 1st case , when the range is 3 km:

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

$\implies\: 3000 = \dfrac{ {u}^{2} \sin(2 \times {15}^{ \circ} ) }{g}$

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

$\implies\: 3000 = \dfrac{ {u}^{2} \times \frac{1}{2} }{g}$

$\implies\: \dfrac{ {u}^{2}}{g} = 3000 \times 2$

$\implies\: \dfrac{ {u}^{2}}{g} = 6000 \: \: \: \: \: ....(1)$

Now , let's calculate the max range , which is obtained at 45° angle of Projection:

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

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

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

$\implies \: R_{max}= \dfrac{ {u}^{2} \times 1 }{g}$

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

$\implies \: R_{max}= 6000 \: m$

$\implies \: R_{max}= 6\: km$

So, the max range that can be achieved is 6 km.

Hence , the Projectile can never reach 8km range at any angle of Projection.