Question

A gun is able to hit a target $100\,{\text{m}}$ away if it shoots at $30^\circ$. If the target moves $100\,{\text{m}}$ away from its original position, at what angle should the gun shoot?
A. ${\sin ^{ - 1}}\left( {\dfrac{1}{3}} \right)$
B. ${\sin ^{ - 1}}\left( {\dfrac{2}{3}} \right)$
C. $45^\circ$
D. Not possible to shoot the target

Answer 1

Use the formula for the horizontal range of the projectile. This formula gives the relation between the horizontal range of the projectile, velocity of projection, angle of projection and acceleration due to gravity. Rewrite this formula for the initial and final condition of the horizontal range and calculate the new angle of projection.

Formula used:
The horizontal range $R$ of the projectile is given by
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$ …… (1)
Here, $u$ is velocity of projection of the projectile, $\theta$ is angle of projection and $g$ is acceleration due to gravity.

Complete step by step answer:
We have given that for the angle of projection of $30^\circ$ the gun, the maximum horizontal range of the gun is $100\,{\text{m}}$.
${\theta _1} = 30^\circ$
${R_1} = 100\,{\text{m}}$
We have asked to calculate the angle of projection of the bullet from the gun in order to hit a target which moves $100\,{\text{m}}$ away from its initial position and the gun is at the same position.Rewrite equation (1) for the initial horizontal range of the gun.
${R_1} = \dfrac{{{u^2}\sin 2{\theta _1}}}{g}$

Substitute $100\,{\text{m}}$ for ${R_1}$ and $30^\circ$ for ${\theta _1}$ in the above equation.
$100\,{\text{m}} = \dfrac{{{u^2}\sin 2\left( {30^\circ } \right)}}{g}$
$\Rightarrow 100\,{\text{m}} = \dfrac{{{u^2}\sin 60^\circ }}{g}$
$\Rightarrow 100\,{\text{m}} = \dfrac{{{u^2}}}{g}\dfrac{{\sqrt 3 }}{2}$ …… (2)
The new horizontal range for the gun is away from its original position.
${R_2} = {R_1} + 100\,{\text{m}}$
$\Rightarrow {R_2} = 100\,{\text{m}} + 100\,{\text{m}}$
$\Rightarrow {R_2} = 200\,{\text{m}}$
Hence, the new horizontal range for the gun is $200\,{\text{m}}$.

Rewrite equation (1) for the changed horizontal range of the gun.
${R_2} = \dfrac{{{u^2}\sin 2{\theta _2}}}{g}$
Here, ${\theta _2}$ is the new angle of projection of the gun.
Substitute $200\,{\text{m}}$ for ${R_2}$ in the above equation.
$200\,{\text{m}} = \dfrac{{{u^2}\sin 2{\theta _2}}}{g}$ …… (3)
Divide equation (3) by equation (1).
$\dfrac{{200\,{\text{m}}}}{{100\,{\text{m}}}} = \dfrac{{\dfrac{{{u^2}\sin 2{\theta _2}}}{g}}}{{\dfrac{{{u^2}}}{g}\dfrac{{\sqrt 3 }}{2}}}$
$\Rightarrow 2 = \dfrac{{\sin 2{\theta _2}}}{{\dfrac{{\sqrt 3 }}{2}}}$
$\therefore \sin 2{\theta _2} = \sqrt 3$
There is no angle to satisfy the above equation. Therefore, it is not possible to shoot the target at a new position.

Hence, the correct option is D.

Note: The students should be careful while determining the new angle of projection for the new horizontal range of the gun. There is no direct angle which gives the new horizontal range as there is no direct angle whose value is equal to square root of 3. The students should be able to understand this and minimize the time in finding the new angle of projection.