A gun can fire shells with a maximum speed $v_0$. On level ground the greatest horizontal range it can achieve (firing at $45^\circ$) is $R = \dfrac{v_0^2}{g}$. A target lies farther away, a distance $\Delta x$ beyond $R$ — that is, at a horizontal distance $R + \Delta x$ from the gun. Show that this target can still be hit with the same gun by raising the gun to a height of at least $$h = \Delta x\left[1 + \frac{\Delta x}{R}\right].$$ (Approach: place the gun at the top of a tower of height $h$; take the launch point as the origin with $x$ horizontal and $y$ vertically upward. The target then sits at $x = R + \Delta x$ and $y = -h$. Fire at the angle that maximises horizontal reach, $45^\circ$.)

A gun can fire shells with a maximum speed v₀. On level ground the greatest horizontal range it can achieve (f

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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 gun's speed is fixed, so to reach farther than $R$ we give the shell extra fall by raising the gun. Firing at $45^\circ$ (maximum horizontal reach) from a tower of height $h$ , we insert the target coordinates $(R+\Delta x,\,-h)$ into the projectile's trajectory equation. Solving gives exactly $h = \Delta x\,[1+\Delta x/R]$ , and because $45^\circ$ is optimal, no smaller height can work.

Concept

The launch speed cannot exceed $v_0$ , so on flat ground the farthest point is $R = v_0^2/g$ (at $45^\circ$ ). To hit a point beyond $R$ , we raise the launch point by $h$ ; the extra height gives the shell more time in the air, extending its horizontal travel. To need the smallest $h$ , we should launch at the angle that carries the shell farthest horizontally, which is $45^\circ$ .

Setup and steps

Take the muzzle as origin, $x$ horizontal, $y$ upward. Launch at $45^\circ$ with speed $v_0$ :

v_x = v_0\cos 45^\circ = \frac{v_0}{\sqrt2}, \qquad v_y = v_0\sin 45^\circ = \frac{v_0}{\sqrt2}.

The equation of the trajectory is

y = x\tan\theta - \frac{g\,x^2}{2v_0^2\cos^2\theta}.

With $\theta = 45^\circ$ ( $\tan\theta = 1$ , $\cos^2\theta = \tfrac12$ ) and using $v_0^2 = gR$ :

y = x - \frac{g\,x^2}{2v_0^2\cdot \tfrac12} = x - \frac{g\,x^2}{v_0^2} = x - \frac{x^2}{R}.

The target is on the ground a height $h$ below the muzzle, at $x = R+\Delta x$ , $y = -h$ :

-h = (R+\Delta x) - \frac{(R+\Delta x)^2}{R}.

Therefore …

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Concept: Use the Parabola's Vertex Form Instead of Plugging Directly Into the Standard Trajectory Equation

Method: Complete the Square on the $45^\circ$ Trajectory

The direct method substitutes the target's coordinates straight into the standard trajectory formula and simplifies. This method first rewrites the $45^\circ$ trajectory in vertex form (highlighting its peak point explicitly), then reads off the answer as a geometric offset from that peak.

Steps

Write the $45^\circ$ trajectory using $v_0^2=gR$ (from the given $R=v_0^2/g$ ):

y = x - \frac{gx^2}{v_0^2} = x - \frac{x^2}{R}.

Complete the square in $x$ to expose the vertex (peak) explicitly:

y = -\frac1R\left(x^2 - Rx\right) = -\frac1R\left[\left(x-\frac R2\right)^2 - \frac{R^2}{4}\right] = \frac R4 - \frac1R\left(x-\frac R2\right)^2.

This vertex form confirms directly that the peak sits at $\left(\tfrac R2,\ \tfrac R4\right)$ .

Substitute the target's position, $x=R+\Delta x$ , $y=-h$ , into the vertex form:

-h = \frac R4 - \frac1R\left(R+\Delta x - \frac R2\right)^2 = \frac R4 - \frac1R\left(\frac R2+\Delta x\right)^2.

Expand the squared term $\left(\tfrac R2+\Delta x\right)^2 = \tfrac{R^2}4 + R\Delta x + \Delta x^2$ , so:

-h = \frac R4 - \frac R4 - \Delta x - \frac{\Delta x^2}{R} = -\Delta x - \frac{\Delta x^2}{R}.

Flip the sign: $h = \Delta x + \frac{\Delta x^2}{R} = \Delta x\left(1+\frac{\Delta x}{R}\right).$ …

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Concept

Setup and steps

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