Two stones are projected from the top of a cliff h metres high, with the same speed u, so as to hit

Q: Wait, why is the vertical coordinate $-h$ and not $h$? Shouldn't the height be positive?

That is a fantastic question! In physics, the sign of your coordinate depends entirely on your chosen origin. If we place our origin at the launch point (the top of the cliff), then the ground is below us. Therefore, any point on the ground must have a negative vertical coordinate. Using $y = -h$ is just a consistent way of saying 'the ground is $h$ meters below the launch point.' If you chose the ground as the origin, the math would still work, but the signs would flip. Consistency is the key to avoiding errors!

Q: Can we just use energy conservation to solve this? It seems faster.

Energy conservation is a powerful tool, but it only tells us about the *speed* of the stone at the moment of impact. It doesn't tell us *where* it lands or the *angle* at which it hits. To find the range and the angle $\theta$, we need to track the position of the stone over time, which is exactly what the trajectory equation does. Energy conservation is like knowing how fast a car is going, but the trajectory equation is like knowing the entire map of the road.

Q: Why can we divide by $\tan \theta$ in the final steps? What if $\theta = 0$?

You are absolutely right to be cautious! Dividing by a variable is only valid if that variable is non-zero. In our problem, the second stone is projected at an angle $\theta$ to the horizontal. If $\theta$ were $0$, it would be the same as the first stone, which we already accounted for. Since the problem implies two distinct paths, we know $\theta \neq 0$, so $\tan \theta \neq 0$. Therefore, dividing by $\tan \theta$ is perfectly safe and mathematically sound.

Question

Two stones are projected from the top of a cliff h metres high, with the same speed u, so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected at an angle $	heta$ to the horizontal then $	an 	heta$ equals

Options:

(A) $u\sqrt{\frac{2}{gh}}$

(B) $\sqrt{\frac{2u}{gh}}$

(C) $2g\sqrt{\frac{u}{h}}$

(D) $2h\sqrt{\frac{u}{g}}$

Sigma Percentile · Accepted Answer

### The Dance of Two Stones: A Journey Through Projectile Motion

Imagine you are standing on the edge of a sheer cliff, the wind whipping past you, looking down at the ground $h$ meters below. You hold two stones. You are about to perform a feat of physics.

You throw the first stone perfectly horizontally with speed $u$. It doesn't just fall; it traces a graceful parabolic arc until it kisses the ground at a specific spot.

Now, you throw the second stone with the same speed $u$, but this time, you aim it upwards at an angle $	heta$. It soars higher, spends more time in the air, and yet, in a moment of pure mathematical harmony, it lands at the exact same spot as the first.

### Phase 1: The Horizontal Stone

Let us begin with the first stone. Because it is thrown horizontally, its initial vertical velocity is zero. It is purely a victim of gravity.

The time it takes to fall a vertical distance $h$ is governed by the kinematic equation:

$$h = \frac{1}{2}gt^2$$

Solving for time, we get $t = \sqrt{\frac{2h}{g}}$. Since the horizontal velocity $u$ remains constant throughout the flight, the horizontal range $R$ is simply $R = u 	imes t$.

Thus, we find our first crucial relationship:

$$R = u \sqrt{\frac{2h}{g}}$$

If we square this, we get:

$$R^2 = \frac{2hu^2}{g}$$

Keep this in your mind; it is the key that will unlock the entire problem.

### Phase 2: The Angled Stone

Now, consider the second stone. It is launched at an angle $	heta$. Its path is governed by the trajectory equation:

$$y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$$

If we set our origin at the launch point, the landing spot is at coordinates $(R, -h)$. We use $-h$ because the ground is below our launch point.

Substituting these coordinates into our trajectory equation, we get:

$$-h = R 	an 	heta - \frac{gR^2}{2u^2 \cos^2 	heta}$$

### Phase 3: The Mathematical Bridge

Here is where the magic happens. Look at the term $\frac{gR^2}{2u^2}$ in our equation.

From our first stone, we know that $R^2 = \frac{2hu^2}{g}$, which implies:

$$\frac{gR^2}{2u^2} = h$$

We can substitute this directly into our trajectory equation. The equation transforms into:

$$-h = R 	an 	heta - \frac{h}{\cos^2 	heta}$$

### Phase 4: The Elegant Simplification

Now, let us rearrange the terms to isolate our target, $	an 	heta$. Moving the terms around, we get:

$$R 	an 	heta = \frac{h}{\cos^2 	heta} - h$$

Factoring out $h$, we have:

$$R 	an 	heta = h \left( \frac{1}{\cos^2 	heta} - 1 ight)$$

Recall your trigonometric identities: $\frac{1}{\cos^2 	heta}$ is simply $\sec^2 	heta$, and $\sec^2 	heta - 1 = 	an^2 	heta$. The equation becomes:

$$R 	an 	heta = h 	an^2 	heta$$

Since $	heta 
eq 0$, we can divide both sides by $	an 	heta$ to get:

$$R = h 	an 	heta$$

### The Final Reveal

We are almost there. We know $R = u \sqrt{\frac{2h}{g}}$. Substituting this back into our simplified equation, we get:

$$u \sqrt{\frac{2h}{g}} = h 	an 	heta$$

Solving for $	an 	heta$, we find:

$$	an 	heta = \frac{u \sqrt{\frac{2h}{g}}}{h}$$

Bringing the $h$ inside the square root as $h^2$, we get $	an 	heta = u \sqrt{\frac{2h}{gh^2}}$, which simplifies to:

$$	an 	heta = u \sqrt{\frac{2}{gh}}$$

And there it is! The answer emerges from the algebra, clean and elegant. You have just mastered the physics of two stones, proving that even the most complex motions are governed by simple, beautiful laws.

Select Answer:

Visualized Solution

Visualizing the Setup

Trajectory of Stone 1

Calculating Range $R$

Trajectory of Stone 2

Equation of Trajectory

Applying Boundary Conditions

Eliminating $u$ and $g$

Algebraic Simplification

Applying Trigonometry

Isolating $tan θ$

Final Answer

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Dance of Two Stones

Phase 1

Phase 2

Phase 3

Phase 4

The Final Reveal

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