A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800m from the foot of hill and can be moved on the ground at a speed of 2 m/s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill ? Take $g = 10$ m/s$^2$.

> [!TLDR] > The shortest time is obtained by NOT moving the cannon at all — fired from its original position, $800\ \text{m}$ from the hill, at the smallest angle that just clears the $500\ \text{m}$ hilltop, the packet reaches the far side in about $20\ \text{s}$. ## What we are minimizing The total time has two parts: the time spent repositioning the cannon (it moves at $2\ \text{m/s}$) and the time of flight of the packet (launched at $125\ \text{m/s}$, $g = 10\ \text{m/s}^2$). Both depend on the launch angle chosen, so different strategies must be compared. ## A tempting but wasteful strategy: put the apex exactly over the hilltop One way to clear the hill is to aim so that the very peak of the trajectory sits exactly at the hilltop's height, $500\ \text{m}$. - Vertical launch speed needed to just reach height $500\ \text{m}$ at the peak: using $v_y^2 = 2gh$, $$v_y = \sqrt{2 \times 10 \times 500} = 100\ \text{m/s}$$ - Horizontal launch speed (from $v_0 = 125\ \text{m/s}$): $v_x = \sqrt{125^2 - 100^2} = \sqrt{5625} = 75\ \text{m/s}$. - Time to reach the peak: $t_1 = \dfrac{v_y}{g} = \dfrac{100}{10} = 10\ \text{s}$, covering a horizontal distance $x = v_x t_1 = 75 \times 10 = 750\ \text{m}$. - So the cannon would have to be moved from $800\ \text{m}$ to $750\ \text{m}$ — a $50\ \text{m}$ move, costing $\dfrac{50}{2} = 25\ \text{s}$. - By symmetry the packet also takes $10\ \text{s}$ to come down the far side, so total flight time $= 20\ \text{s}$. - **Total time this way: $25 + 20 = 45\ \text{s}$.** This clears the hill, but it spends $25\ \text{s}$ of repositioning just to place the peak exactly over the hilltop — far more than is actually needed, since the packet only has to *clear* the hill, not peak exactly above it. ## The better strategy: don't move the cannon at all Leave the cannon at its original $800\ \text{m}$ and fire it at the smallest angle $\theta$ that still lets the packet be at height $500\ \text{m}$ (or higher) when it is horizontally $8

A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed o

NCERT Exemplar · Q29

g = 10

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Concept understanding — Projectile Motion

Projectile Motion — From Intuition to Precision

Imagine you throw a ball to a friend. It doesn't travel in a straight line — it rises, slows down, then curves downward and falls. That curved path is a projectile's trajectory. The ball is a projectile: any object that is launched into the air and then moves only under the influence of gravity (and air resistance, which we ignore for now).

The key intuition: once the ball leaves your hand, the only force acting on it is gravity pulling it straight down. There is no forward force after release. The ball keeps moving forward because of inertia — it wants to keep going in a straight line at constant speed. But gravity keeps pulling it down, so the forward motion and downward acceleration combine to produce a curved path.

The Precise Statement

g \approx 9.8 \, \text{m/s}^2

We break the motion into two independent components:

Horizontal motion: No acceleration ( $a_x = 0$ ). So horizontal velocity $v_x$ is constant.
Vertical motion: Constant downward acceleration ( $a_y = -g$ ). So vertical velocity $v_y$ changes linearly with time.

The independence of these components is the central idea — what happens vertically does not affect what happens horizontally, and vice versa.

The Equations (for a projectile launched with initial speed $u$ at angle $\theta$ above horizontal)

First, resolve the initial velocity:

u_x = u \cos\theta, \quad u_y = u \sin\theta

Horizontal motion (constant velocity):

x = u_x t = (u \cos\theta) t

Vertical motion (constant acceleration $-g$ ):

v_y = u_y - gt = u \sin\theta - gt

y = u_y t - \frac{1}{2} g t^2 = (u \sin\theta) t - \frac{1}{2} g t^2

Key Results You Must Know

Important

Time of flight $T$ : total time the projectile stays in the air (until $y=0$ again).

$T = \frac{2u \sin\theta}{g}$

Important

Maximum height $H$ : the highest vertical position reached (when $v_y = 0$ ).

$H = \frac{u^2 \sin^2\theta}{2g}$

Important

Range $R$ : the horizontal distance covered when it returns to launch height.

$R = \frac{u^2 \sin 2\theta}{g}$

Note

The range is maximum when $\sin 2\theta = 1$ , i.e., $\theta = 45^\circ$ . For a given speed, $45^\circ$ gives the farthest throw.

The Trajectory Equation (Path Shape)

Eliminate $t$ from the $x$ and $y$ equations to get $y$ as a function of $x$ :

y = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta}

This is a parabola — the signature shape of projectile motion.

Common Mistake to Avoid …

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The shortest time is obtained by NOT moving the cannon at all — fired from its original position, $800\ \text{m}$ from the hill, at the smallest angle that just clears the $500\ \text{m}$ hilltop, the packet reaches the far side in about $20\ \text{s}$ .

What we are minimizing

The total time has two parts: the time spent repositioning the cannon (it moves at $2\ \text{m/s}$ ) and the time of flight of the packet (launched at $125\ \text{m/s}$ , $g = 10\ \text{m/s}^2$ ). Both depend on the launch angle chosen, so different strategies must be compared.

A tempting but wasteful strategy: put the apex exactly over the hilltop

One way to clear the hill is to aim so that the very peak of the trajectory sits exactly at the hilltop's height, $500\ \text{m}$ .

Vertical launch speed needed to just reach height $500\ \text{m}$ at the peak: using $v_y^2 = 2gh$ ,

v_y = \sqrt{2 \times 10 \times 500} = 100\ \text{m/s}

Horizontal launch speed (from $v_0 = 125\ \text{m/s}$ ): $v_x = \sqrt{125^2 - 100^2} = \sqrt{5625} = 75\ \text{m/s}$ .
Time to reach the peak: $t_1 = \dfrac{v_y}{g} = \dfrac{100}{10} = 10\ \text{s}$ , covering a horizontal distance $x = v_x t_1 = 75 \times 10 = 750\ \text{m}$ .
So the cannon would have to be moved from $800\ \text{m}$ to $750\ \text{m}$ — a $50\ \text{m}$ move, costing $\dfrac{50}{2} = 25\ \text{s}$ .
By symmetry the packet also takes $10\ \text{s}$ to come down the far side, so total flight time $= 20\ \text{s}$ .
Total time this way: $25 + 20 = 45\ \text{s}$ .

This clears the hill, but it spends $25\ \text{s}$ of repositioning just to place the peak exactly over the hilltop — far more than is actually needed, since the packet only has to clear the hill, not peak exactly above it.

The better strategy: don't move the cannon at all

Leave the cannon at its original $800\ \text{m}$ and fire it at the smallest angle $\theta$ that still lets the packet be at height $500\ \text{m}$ (or higher) when it is horizontally $800\ \text{m}$ away — i.e. just grazing the hilltop, not peaking over it.

The trajectory height at horizontal distance $x$ is:

y = x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta} = x\tan\theta - \frac{gx^2}{2v_0^2}\sec^2\theta

At $x = 800\ \text{m}$ , $v_0 = 125\ \text{m/s}$ , $g = 10\ \text{m/s}^2$ :

\frac{gx^2}{2v_0^2} = \frac{10 \times 800^2}{2 \times 125^2} = \frac{6{,}400{,}000}{31{,}250} = 204.8

Setting $y = 500\ \text{m}$ (just grazing the hilltop) and writing $t = \tan\theta$ , with $\sec^2\theta = 1 + t^2$ :

800t - 204.8(1+t^2) = 500 \implies 204.8t^2 - 800t + 704.8 = 0

Solving this quadratic:

$t = \frac{800 \pm \sqrt{800^2 - 4(204.8)(704.8)}}{2(204.8)} \approx \frac{800 \pm 250.6}{409.6}$ …

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Concept: Show the Optimum Is a Boundary Point via Sensitivity, Not by Comparing Two Guessed Strategies

Method: Perturb the Cannon's Position and Compare Marginal Cost to Marginal Benefit

Rather than picking two specific candidate cannon positions and comparing their total times, this method asks how the total time changes if the cannon is nudged closer to the hill by a small amount, and shows the cost of moving always outweighs any benefit -- proving $x=800\text{ m}$ (no move at all) is optimal without needing to guess the right alternative in advance.

Steps

Write the total time as a function of the cannon's distance $x$ from the hill ( $x\le800\text{ m}$ , moved at $2\text{ m/s}$ ):

T(x) = \underbrace{\frac{800-x}{2}}_{\text{repositioning cost}} + \underbrace{\frac{2v_0\sin\theta_{\min}(x)}{g}}_{\text{flight time}},

where $\theta_{\min}(x)$ is the smallest launch angle that still clears the $500\text{ m}$ hill from distance $x$ (found from $x\tan\theta-\tfrac{gx^2}{2v_0^2}\sec^2\theta=500$ ).

The repositioning term has a fixed, unavoidable marginal cost: moving the cannon $1\text{ m}$ closer always costs $\tfrac12\text{ s}$ , regardless of $x$ .
Check whether the flight-time term compensates, using a second data point close to $x=800\text{ m}$ . Solve the same quadratic at $x=700\text{ m}$ (cannon moved $100\text{ m}$ closer):

156.8\tan^2\theta - 700\tan\theta + 656.8 = 0 \;\Rightarrow\; \tan\theta_{\min}\approx1.3415 \;\Rightarrow\; \theta_{\min}\approx53.3^\circ.

This is essentially identical to the $\theta_{\min}\approx53.3^\circ$ found at $x=800\text{ m}$ -- so the flight-time term $2v_0\sin\theta_{\min}/g\approx20.0\text{ s}$ barely changes at all between the two positions.

Compare the two effects. Moving $100\text{ m}$ closer costs a guaranteed $50\text{ s}$ in repositioning, but buys essentially zero reduction in flight time (both $\approx20.0\text{ s}$ ) -- so the total time strictly increases the moment the cannon is moved. …

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What we are minimizing

A tempting but wasteful strategy: put the apex exactly over the hilltop

The better strategy: don't move the cannon at all

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