UPSC 2026 Maths Optional Paper 1 Q6b-i — Step-by-Step Solution

10 marks · Section B

Projectile motion · Dynamics & Statics · asked 6× in 14 yrs · Read the full method →

Question

A battleship is steaming ahead with velocity $u$ . A gun is mounted on the ship so as to point straight backwards and is set at an angle of elevation $\alpha$ . If $v$ be the velocity of projection relative to the gun, show that the range is $\dfrac{2v}{g}\sin\alpha\,(v\cos\alpha - u)$ and the angle for maximum range is

\cos^{-1}\left(\frac{u + \sqrt{u^2 + 8v^2}}{4v}\right)

where $g$ stands for acceleration due to gravity.

Technique

Relative-velocity composition to get the projectile’s absolute velocity (the gun points backwards, so the ship’s speed subtracts from the horizontal component), then standard projectile range on a horizontal plane; finally maximise the range over $\alpha$ , which reduces to a quadratic in $\cos\alpha$ .

Solution

Step 1 — Set up the geometry and get the absolute velocity of projection.

Work in the vertical plane containing the ship’s line of motion. Let the ship steam with speed $u$ in the direction $\overrightarrow{OF}$ (“forward”), and take the backward horizontal direction $\overrightarrow{OB}$ (opposite to the ship’s motion) as the positive $x$ -axis, with the upward vertical as the positive $y$ -axis. Let $O$ be the muzzle at the instant of firing.

The muzzle imparts to the shot a velocity $v$ relative to the gun, directed backwards and inclined at $\alpha$ to the horizontal; its components along $(\hat x,\hat y)$ are $(v\cos\alpha,\;v\sin\alpha)$ .

The gun itself (with the ship) is moving with velocity $u$ forward, i.e. $(-u,\,0)$ in our axes.

By the triangle of velocities, the velocity of the shot relative to the ground (the water) is

\vec V=\underbrace{(v\cos\alpha,\;v\sin\alpha)}_{\text{relative to gun}}+\underbrace{(-u,\;0)}_{\text{of gun}}=\bigl(v\cos\alpha-u,\;\;v\sin\alpha\bigr).

So the shot leaves with horizontal component $v\cos\alpha-u$ (backwards) and vertical component $v\sin\alpha$ . Note the vertical component is unaffected by the ship’s motion, since that motion is horizontal.

Step 2 — Projectile motion and the range.

After projection the only force is gravity, so the horizontal component stays constant and the vertical motion is uniformly retarded:

x=(v\cos\alpha-u)\,t,\qquad y=(v\sin\alpha)\,t-\tfrac12 g t^{2}.

The shot returns to the horizontal plane of projection when $y=0$ , $t\neq0$ :

T=\frac{2v\sin\alpha}{g}\qquad(\text{time of flight}).

Hence the range (measured on the water, from the point of projection) is

R=(v\cos\alpha-u)\,T=\frac{2v}{g}\sin\alpha\,(v\cos\alpha-u).\qquad\blacksquare

Step 3 — Maximise $R$ over $\alpha$ .

Since $\dfrac{2v}{g}>0$ is a constant, maximise

f(\alpha)=\sin\alpha\,(v\cos\alpha-u)=\tfrac12 v\sin2\alpha-u\sin\alpha,\qquad 0<\alpha<\tfrac{\pi}{2}.

Differentiate:

f'(\alpha)=v\cos2\alpha-u\cos\alpha=v\bigl(2\cos^{2}\alpha-1\bigr)-u\cos\alpha .

Setting $f'(\alpha)=0$ and writing $c=\cos\alpha$ gives the quadratic in $\cos\alpha$

2v\,c^{2}-u\,c-v=0\quad\Longrightarrow\quad c=\frac{u\pm\sqrt{u^{2}+8v^{2}}}{4v}.

Step 4 — Select the root and confirm it is a maximum.

The root with the minus sign is negative (since $\sqrt{u^{2}+8v^{2}}>u$ ), giving $\alpha>\dfrac{\pi}{2}$ — the gun would be pointing below the horizontal in the backward sense, which is not an elevation. So take the plus sign:

\cos\alpha=\frac{u+\sqrt{u^{2}+8v^{2}}}{4v}.

For the second-derivative test,

f''(\alpha)=-2v\sin2\alpha+u\sin\alpha=-\sin\alpha\,\bigl(4v\cos\alpha-u\bigr).

At the chosen root $4v\cos\alpha=u+\sqrt{u^{2}+8v^{2}}>u$ , so $4v\cos\alpha-u>0$ ; and $\sin\alpha>0$ on $(0,\tfrac\pi2)$ . Hence $f''<0$ and the stationary point is a maximum. (At the rejected root $4v\cos\alpha-u=-\sqrt{u^{2}+8v^{2}}<0$ , so it is a minimum — a clean confirmation that the sign choice is forced, not cosmetic.)

Therefore

\alpha_{\max}=\cos^{-1}\!\left(\frac{u+\sqrt{u^{2}+8v^{2}}}{4v}\right).\qquad\blacksquare

Step 5 — Admissibility.

The value is a genuine elevation only if $\cos\alpha\le1$ :

u+\sqrt{u^{2}+8v^{2}}\le 4v\iff \sqrt{u^{2}+8v^{2}}\le 4v-u\iff u^{2}+8v^{2}\le16v^{2}-8uv+u^{2}\iff u\le v .

The same condition makes the range positive: $v\cos\alpha>u\iff\sqrt{u^{2}+8v^{2}}>3u\iff v>u$ . So for $u<v$ (the physically intended case) the formula gives an admissible angle and a positive maximum range; if $u\ge v$ the ship outruns the shot’s backward horizontal component and no positive backward range exists at any elevation.

Answer

\boxed{\;R=\frac{2v}{g}\sin\alpha\,(v\cos\alpha-u),\qquad \alpha_{\max}=\cos^{-1}\!\left(\frac{u+\sqrt{u^{2}+8v^{2}}}{4v}\right)\;(u<v).\;}