Q.Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of $45^\circ$ by equal amounts, the ranges are equal”. Prove this statement.

Concept and Intuition

Projectile range depends on launch speed and angle through $R=\dfrac{u^2\sin2\theta}{g}$. Two different angles can give the same range only if the speeds compensate for the difference in $\sin2\theta$ — this is the entire content of the problem.

Step-by-Step Solution

1. Range formula: $R=\dfrac{u^2\sin2\theta}{g}$.
2. For $\theta_1=15^\circ$: $R_1=\dfrac{u_1^2\sin30^\circ}{g}$.
3. For $\theta_2=45^\circ$: $R_2=\dfrac{u_2^2\sin90^\circ}{g}$.
4. Given $R_1=R_2$: $u_1^2\sin30^\circ=u_2^2\sin90^\circ$. …

Concept and Intuition

For a fixed launch speed, range $R=\dfrac{u^2\sin2\theta}{g}$ is maximum when $\sin2\theta=1$, i.e. $\theta=45^\circ$. During flight, the horizontal velocity component is constant while the vertical component shrinks to zero at the top and grows again — so the speed (magnitude of velocity) is smallest exactly at the highest point of the trajectory.

Step-by-Step Solution

1. "Maximum possible range" $\Rightarrow \theta=45^\circ$ and $R_{max}=\dfrac{u^2}{g}=200$ m.
2. At $45^\circ$: maximum height $H=\dfrac{u^2\sin^2\theta}{2g}=\dfrac{u^2}{4g}=\dfrac{R}{4}=\dfrac{200}{4}=50$ m.
3. The apex occurs at the midpoint of the range: horizontal distance $=\dfrac{R}{2}=100$ m. …

Concept and Intuition

For a projectile launched with speed $u$ at angle $\alpha$ to the horizontal, only the vertical component of the initial velocity, $u\sin\alpha$, governs how high it rises — the horizontal component doesn't affect height at all. Maximum height is:

Two projectiles launched at complementary angles $\theta$ and $90^\circ-\theta$ with the same speed have the same range (a classic result), but very different heights, because $\sin\theta \ne \sin(90^\circ-\theta)$ in general.

Step-by-Step Solution

1. Both A and B are launched with the same speed $u$.
2. $H_A = \dfrac{u^2\sin^2\theta}{2g}$ (angle $\theta$). …

Concept and Intuition

For a fixed launch speed, the range $R=\dfrac{V^2\sin2\theta}{g}$ is the same for angles $\theta$ and $90°-\theta$, since $\sin(2(90°-\theta))=\sin(180°-2\theta)=\sin2\theta$. This is the classic "complementary angles give equal range" result.

Step-by-Step Solution

1. Let one body be launched at angle $\theta$ and the other at $90°-\theta$ (both give the same range for the same speed $V$).
2. Time of flight formula: $T=\dfrac{2V\sin(\text{angle})}{g}$.
3. So $T_1=\dfrac{2V\sin\theta}{g}$ and $T_2=\dfrac{2V\sin(90°-\theta)}{g}=\dfrac{2V\cos\theta}{g}$.
4. Ratio: …

Concept and Intuition

The horizontal range of a projectile launched at angle $\theta$ with speed $u$ is $R=\frac{u^2\sin2\theta}{g}$. Since $\sin2\theta=\sin(180°-2\theta)$, any given range (except the maximum at $45°$) is achieved by two complementary angles $\theta$ and $90°-\theta$ — this is why the answer is naturally a pair of angles.

Step-by-Step Solution

1. Range formula: $R=\dfrac{u^2\sin2\theta}{g}$.
2. Substitute $u=60$, $R=180\sqrt3$, $g=10$:

3. Solve: $\sin2\theta=\dfrac{180\sqrt3}{360}=\dfrac{\sqrt3}{2}$. …

Concept and Intuition

Range of a projectile depends on $\sin(2\theta)$, which is symmetric about $45^\circ$ — a projectile fired at $\theta$ and one fired at $90^\circ-\theta$ (same speed) land at the same spot, but one is a 'flat' shot and the other a 'lofted' shot with a longer time of flight. Time of flight, unlike range, depends on $\sin\theta$ alone (not $\sin2\theta$), so it does NOT share the same symmetry — this is exactly why the two times differ even though the ranges are equal.

Step-by-Step Solution

1. Range: $R = \dfrac{u^2\sin(2\theta)}{g}$. For bodies A and B (same $u$), equal range means $\sin(2\theta_1) = \sin(2\theta_2)$.
2. This is satisfied (for two distinct angles) when $2\theta_2 = 180^\circ - 2\theta_1$, i.e. $\theta_2 = 90^\circ - \theta_1$ — the two angles are complementary.
3. Time of flight: $T = \dfrac{2u\sin\theta}{g}$. …

Concept and Intuition

For a fixed launch speed $u$, the range $R=\dfrac{u^2\sin2\theta}{g}$ is the same for $\theta$ and $90^\circ-\theta$ (since $\sin2\theta=\sin(180^\circ-2\theta)$). This symmetry connects the two times of flight neatly to the common range.

Step-by-Step Solution

1. Let the two complementary angles be $\theta$ and $90^\circ-\theta$, both giving range $R=\dfrac{u^2\sin2\theta}{g}$.
2. Time of flight for angle $\theta$: $T_1=\dfrac{2u\sin\theta}{g}$. For angle $90^\circ-\theta$: $T_2=\dfrac{2u\sin(90^\circ-\theta)}{g}=\dfrac{2u\cos\theta}{g}$.
3. Product: $T_1T_2=\dfrac{4u^2\sin\theta\cos\theta}{g^2}=\dfrac{2u^2\sin2\theta}{g^2}$.
4. Since $R=\dfrac{u^2\sin2\theta}{g}\Rightarrow u^2\sin2\theta=Rg$, substitute: $T_1T_2=\dfrac{2Rg}{g^2}=\dfrac{2R}{g}$. …

Concept and Intuition

Two projectiles launched at complementary angles $\theta$ and $90^\circ-\theta$ with the same speed have the same range, but generally different maximum heights (unless $\theta=45^\circ$). The height formula $H=\dfrac{u^2\sin^2(\text{angle})}{2g}$ lets us directly compare the two heights.

Step-by-Step Solution

1. Height for angle $\theta$: $H_1=\dfrac{u^2\sin^2\theta}{2g}$.
2. Height for angle $90^\circ-\theta$: $H_2=\dfrac{u^2\sin^2(90^\circ-\theta)}{2g}=\dfrac{u^2\cos^2\theta}{2g}$.
3. Given $H_2-H_1=10$ m (second stone rises 10 m higher): $\dfrac{u^2}{2g}(\cos^2\theta-\sin^2\theta)=10$.
4. Using $\cos^2\theta-\sin^2\theta=\cos2\theta$: $\dfrac{u^2}{2g}\cos2\theta=10$. …

Concept and Intuition

For a projectile launched with speed $u$ at angle $\theta$, the range is $R(\theta) = \dfrac{u^{2}\sin 2\theta}{g}$, maximized when $\sin 2\theta = 1$, i.e. $\theta = 45^{\circ}$, giving $R_{\max} = u^{2}/g$. The maximum height for a given launch is $H = \dfrac{u^{2}\sin^{2}\theta}{2g}$. Since the maximum-range condition fixes $\theta = 45^{\circ}$, we substitute that specific angle to find the corresponding height, then express it in terms of $R$ using the range relation.

Step-by-Step Solution

1. Note that "maximum range" implies $\theta = 45^{\circ}$, and at this angle $R = \dfrac{u^{2}}{g}$, so $u^{2} = Rg$.
2. Compute the height at $45^{\circ}$: $H = \dfrac{u^{2}\sin^{2}45^{\circ}}{2g} = \dfrac{u^{2}(1/2)}{2g} = \dfrac{u^{2}}{4g}$. …

Concept and Intuition

When a projectile is launched at angle $\theta$ or its complement $90°-\theta$ with the same speed, both give the same horizontal range (since $\sin2\theta=\sin(2(90°-\theta))$), but different maximum heights. This question exploits the relationship between those two heights and the common range — a well-known projectile motion identity worth memorizing.

Step-by-Step Solution

1. Maximum height for angle $\theta$:

2. Maximum height for angle $(90°-\theta)$, using $\sin(90°-\theta)=\cos\theta$:

3. Multiply them:

Take the square root:

4. The horizontal range for angle $\theta$ (which equals the range for $90°-\theta$ too): …

Concept and Intuition

The horizontal range of a projectile launched with speed $u$ at angle $\theta$ to the horizontal is $R=\frac{u^2\sin2\theta}{g}$. Two projectiles launched from the same point can have equal ranges even with very different speeds and angles, as long as their $u^2\sin2\theta$ products match. Here we're told B's speed and angle explicitly, so we can compute B's range fully, then set A's range formula (with A's unknown angle) equal to it and solve.

Step-by-Step Solution

1. Range of projectile B (speed $0.5v$, angle $45°$): $R_B = \dfrac{(0.5v)^2\sin(2\times45°)}{g} = \dfrac{0.25v^2 \sin90°}{g} = \dfrac{0.25v^2}{g}$ (since $\sin90°=1$).
2. Range of projectile A (speed $v$, unknown angle $\theta_A$): $R_A = \dfrac{v^2\sin2\theta_A}{g}$.
3. Given ranges are equal, $R_A = R_B$: $\dfrac{v^2\sin2\theta_A}{g} = \dfrac{0.25v^2}{g}$. …

Concept and Intuition: Projectile Range Symmetry on an Incline

When two particles start from the same point on an inclined plane and meet again on that plane, their displacements along the incline must be equal at the same time. Particle $A$ simply slides down under gravity (no initial velocity), while particle $B$ is projected at an angle $\beta$ above the horizontal. The trick is to treat the motion along the incline and perpendicular to it separately. For $B$, the component of gravity along the incline is $g \sin \alpha$, and perpendicular to it is $g \cos \alpha$. The condition for meeting is that both particles have the same downhill displacement at the same instant. This yields a relationship between $\alpha$ and $\beta$ that is independent of $u$ — a beautiful symmetry.

Step-by-step solution

1. Set up coordinate axes along the incline.

   Let the $x$-axis point down the incline (the direction of motion for both particles) and the $y$-axis be perpendicular to the incline (pointing upward away from the surface). The acceleration due to gravity $g$ has components:

   • Along the incline: $g_x = g \sin \alpha$ (downhill, positive)
   • Perpendicular to the incline: $g_y = -g \cos \alpha$ (negative because it points into the plane)

2. Motion of particle $A$ (released from rest).

   Particle $A$ starts from $P$ with zero initial velocity and accelerates down the incline. Its displacement along the incline after time $t$ is:

3. Motion of particle $B$ (projected with speed $u$ at angle $\beta$ to the horizontal).

   The initial velocity $u$ makes an angle $\beta$ with the horizontal. The incline itself is at angle $\alpha$ to the horizontal, so the angle between $u$ and the incline is $\beta - \alpha$ (since both are measured from the horizontal, and the incline is rotated by $\alpha$). Therefore, the components of $u$ along and perpendicular to the incline are:

   • Along the incline: $u_x = u \cos(\beta - \alpha)$ (positive, downhill)
   • Perpendicular to the incline: $u_y = u \sin(\beta - \alpha)$ (positive, upward away from the surface)

   The acceleration components for $B$ are the same as for $A$: $a_x = g \sin \alpha$ (downhill) and $a_y = -g \cos \alpha$ (into the incline).

   The displacement of $B$ along the incline after time $t$ is:

4. Condition for meeting on the incline. Both particles start at $P$ at $t=0$ and meet again on the incline at some later time $t = T$. At that moment, their displacements along the incline are equal:

Substituting:

The term $\frac{1}{2} g \sin \alpha \, T^2$ cancels from both sides, leaving:

Since $u \neq 0$ and $T \neq 0$ (they meet after some time), this would imply $\cos(\beta - \alpha) = 0$, which is not generally true. Wait — this cancellation suggests that the meeting condition is not simply equal displacements at the same time? Let's re-examine.

Correction: The above cancellation shows that if both particles have the same acceleration along the incline, then the only way for their displacements to be equal at the same time is if the initial velocity term vanishes — which is impossible unless $B$ is also released from rest. This means we have missed something: Particle $B$ does not stay on the incline; it is a projectile that leaves the surface and then lands back on it. The meeting point is where $B$ returns to the incline after its flight. So the time $T$ is the time of flight of $B$ until it hits the incline again. For $A$, it simply slides down the incline, so its displacement at time $T$ is given by the same formula. But the key is that $B$'s motion perpendicular to the incline determines $T$.

5. Find the time of flight $T$ for particle $B$ to return to the incline. Consider the motion perpendicular to the incline. The initial perpendicular velocity is $u_y = u \sin(\beta - \alpha)$, and the perpendicular acceleration is $a_y = -g \cos \alpha$. The displacement perpendicular to the incline at time $t$ is:

The particle starts on the incline ($y=0$ at $t=0$) and returns to the incline when $y(T)=0$ again (for $T>0$). So:

Factor $T$ (non-zero):

Hence:

6. Now equate the displacements along the incline at time $T$. For particle $A$:

For particle $B$:

Setting $s_A = s_B$:

Cancel the common term:

Again we get the same apparent contradiction. But wait — this is actually correct! It tells us that for the displacements to be equal, we must have $\cos(\beta - \alpha) = 0$, i.e., $\beta - \alpha = \frac{\pi}{2}$. That would mean $\beta = \alpha + \frac{\pi}{2}$, which is impossible because $\beta$ is an angle above the horizontal and $\alpha$ is less than $\frac{\pi}{2}$. So what's wrong?

The subtlety: Particle $A$ is released from $P$ and slides down. Particle $B$ is projected from $P$ and follows a projectile path. They meet again on the incline. But note: $B$'s initial velocity has a component up the incline if $\beta > \alpha$, or down the incline if $\beta < \alpha$. For them to meet, $B$ must be projected in such a way that it goes up and then comes back down, or it goes directly down but with a different time. The equation $s_A = s_B$ gave $0 = u \cos(\beta - \alpha) T$, which forces $\cos(\beta - \alpha)=0$ — this is the condition for $B$ to have zero initial velocity along the incline. That would mean $B$ is projected perpendicular to the incline. But then $B$ would go straight up and down, landing back at $P$, not meeting $A$ elsewhere. So our assumption that they meet at the same time with equal displacements is correct, but we must realize that the meeting point is not necessarily at the same instant as $B$'s return to the incline? No, they start simultaneously and meet again, so the time is the same.

Let's re-read the problem: "If both $A$ and $B$ meet again on the inclined plane". This means they start at $P$ at $t=0$ and later are at the same point on the incline. For $A$, it only moves down the incline. For $B$, it is projected at angle $\beta$ to the horizontal. The only way they can meet is if $B$ is projected downward along the incline? But the figure shows $\beta$ above the horizontal, and the incline slopes downward from $P$. So $\beta$ is measured from the horizontal, and the incline is at angle $\alpha$ below the horizontal from $P$. So the direction of the incline from $P$ is at an angle $-\alpha$ relative to the horizontal. The projection angle $\beta$ is positive (above horizontal). So $B$ is thrown upward relative to the horizontal, but the incline goes downward. This means $B$ initially goes away from the incline, then arcs back and hits it. That is the typical "projectile on an incline" problem.

The meeting condition is that at time $T$, both have the same position vector from $P$. That gives two equations: one along the incline and one perpendicular. But $A$ always stays on the incline, so its perpendicular displacement is always zero. For $B$, at time $T$, its perpendicular displacement must also be zero (since it lands on the incline). That gives the time of flight $T$ as above. Then the along-incline displacements must match. But we saw that leads to $\cos(\beta - \alpha)=0$. This is a paradox.

Resolution: The error is in the sign of the acceleration for $A$. Particle $A$ is released from rest and slides down. Its acceleration along the incline is $g \sin \alpha$ (positive downhill). But $B$'s acceleration along the incline is also $g \sin \alpha$ (downhill). So both have the same acceleration. If they start at the same point and have the same acceleration, their relative velocity is constant. For them to meet again, the relative velocity must be zero at the start, meaning $B$ must have zero initial velocity along the incline relative to $A$. That is, $u \cos(\beta - \alpha) = 0$. So indeed $\beta - \alpha = \frac{\pi}{2}$. But that is impossible geometrically. So what gives?

Actually, the problem states that $A$ is released (so initial velocity zero) and $B$ is projected with initial velocity $u$ at angle $\beta$ to the horizontal. The figure shows $\beta$ measured from the horizontal, and the incline is at angle $\alpha$ to the horizontal. If $\beta$ is above the horizontal, then $\beta - \alpha$ is the angle of projection relative to the incline. For them to meet, $B$ must be projected directly toward the incline? No, the typical result for two particles on an incline where one is dropped and the other is projected is that they meet if the projection is along the line of the incline. But here the projection is at an angle.

Let's check the classic result: For a particle projected from a point on an incline to land back on the incline, the range along the incline is $R = \frac{2u^2 \sin(\beta - \alpha) \cos \beta}{g \cos^2 \alpha}$. For a particle released from rest, its distance down the incline in time $t$ is $\frac{1}{2} g \sin \alpha t^2$. Equating these for the same $t$ gives a relation. But the time for $B$ to land is $T = \frac{2u \sin(\beta - \alpha)}{g \cos \alpha}$. Substitute into $A$'s distance: $s_A = \frac{1}{2} g \sin \alpha \left( \frac{2u \sin(\beta - \alpha)}{g \cos \alpha} \right)^2 = \frac{2u^2 \sin^2(\beta - \alpha) \sin \alpha}{g \cos^2 \alpha}$. For $B$, the range along the incline is $R = \frac{2u^2 \sin(\beta - \alpha) \cos \beta}{g \cos^2 \alpha}$. Set $s_A = R$:

Cancel common factors ($2u^2/(g \cos^2 \alpha)$ and one $\sin(\beta - \alpha)$):

Use identity: $\cos \beta = \cos((\beta - \alpha) + \alpha) = \cos(\beta - \alpha) \cos \alpha - \sin(\beta - \alpha) \sin \alpha$. Substitute:

Bring terms together:

Divide both sides by $\cos(\beta - \alpha) \cos \alpha$ (assuming non-zero):

So:

This is not one of the simple options. But we can rewrite using tangent addition formula. Alternatively, note that $\tan(\beta - \alpha) = \frac{\tan \beta - \tan \alpha}{1 + \tan \beta \tan \alpha}$. Set equal to $\frac{1}{2 \tan \alpha}$:

Cross-multiply:

This still doesn't simplify nicely. But wait — the classic result for "projectile on an incline" where the projection is up the incline gives a different relation. Perhaps the intended interpretation is that both particles move along the incline? But $B$ is projected at an angle $\beta$ to the horizontal, not necessarily along the incline.

Let's re-read the problem statement carefully: "Particle $A$ is released from a point $P$ on a smooth inclined plane... Simultaneously from $P$ another particle $B$ is projected in the same direction, with an initial velocity $u$, at an angle $\beta$ with the horizontal." The phrase "in the same direction" is ambiguous: same direction as what? Possibly same direction as the incline? But then it says "at an angle $\beta$ with the horizontal", so the direction is specified by $\beta$. The figure shows $\beta$ measured from the horizontal, and the incline is at angle $\alpha$. So "same direction" likely means both are moving down the incline? But $A$ is released (so it goes down the incline), and $B$ is projected "in the same direction" — that would mean $B$ is also projected down the incline, i.e., $\beta = \alpha$? That can't be because then they'd never meet (same acceleration, different initial speeds). So "same direction" probably means both are moving in the same general sense (downward to the right), but $B$ is projected at an angle $\beta$ above the horizontal, which could be above or below the incline.

Given the multiple-choice options are all of the form $\alpha + \beta = \text{constant}$, the answer is likely $\alpha + \beta = \frac{\pi}{2}$. Let's test: if $\beta = \frac{\pi}{2} - \alpha$, then $\tan \beta = \cot \alpha = \frac{1}{\tan \alpha}$. Then from the equation $\tan \alpha \tan \beta = 1 + 2 \tan^2 \alpha$, we get $\tan \alpha \cdot \frac{1}{\tan \alpha} = 1 = 1 + 2 \tan^2 \alpha$, which implies $\tan \alpha = 0$, impossible. So that's not it.

Let's derive properly from the condition that the time of flight of $B$ equals the time for $A$ to slide to the meeting point. Let $T$ be the time when they meet. For $A$: $s = \frac{1}{2} g \sin \alpha \, T^2$. For $B$: the time of flight to return to the incline is $T = \frac{2u \sin(\beta - \alpha)}{g \cos \alpha}$. The range of $B$ along the incline is $R = \frac{2u^2 \sin(\beta - \alpha) \cos \beta}{g \cos^2 \alpha}$. Set $s = R$: