A student throws large number of small pebbles in all possible directions with equal speeds u out of

Q: Why did we square the $\tan\phi$ equation?

Squaring $\tan\phi$ removes the square root from the vertical velocity term, making the algebraic manipulation and differentiation (or logical minimization) much simpler.

Q: Why is the minimum angle of impact corresponding to horizontal projection?

When thrown horizontally ($\alpha = 0$), the pebble has the maximum possible horizontal velocity at the ground (since $v_x = u \cos\alpha$). A larger horizontal velocity component makes the resultant velocity vector less steep, thus minimizing the angle with the ground.

Q: Could the minimum angle occur if we throw the pebble downwards?

No. If you throw it downwards ($\alpha < 0$), the vertical velocity at the ground increases, and the horizontal velocity decreases compared to horizontal projection. Both these effects increase the angle of impact.

Q: What does 'angle $\theta$ or greater' mean physically?

It means that out of all the pebbles thrown in every possible direction, the one that hits the ground with the shallowest (flattest) trajectory makes an angle $\theta$. All other pebbles hit at steeper angles.

Q: Why doesn't the mass of the pebble matter?

Because we are neglecting air resistance. In a vacuum, all objects fall with the same acceleration $g$, regardless of their mass. The kinematic equations are independent of mass.

Question

A student throws large number of small pebbles in all possible directions with equal speeds $u$ out of a window. The pebbles hit the horizontal ground moving at an angle $	heta$ or greater with the ground. Air resistance is negligible and acceleration due to gravity is $g$. Deduce suitable expression for the height of the point of projection above the ground.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at a window, a height $h$ above the ground. You have a handful of pebbles, and you start throwing them in every conceivable direction—up, down, horizontally, and at every angle in between.

The only constant is that every pebble leaves your hand with the exact same initial speed, $u$.

As these pebbles trace their parabolic paths and eventually strike the ground, they will hit at various angles. The problem gives us a fascinating constraint: the pebbles hit the ground moving at an angle $	heta$ **or greater**.

This implies that $	heta$ is the **absolute minimum angle of impact**. Our mission is to find the height $h$ that makes this true.

### The Master Equation

To understand the angle of impact, we need to look at the velocity of a pebble just as it kisses the ground.

Let's call the angle it makes with the horizontal $\phi$. From basic vector geometry, the tangent of this angle is the ratio of its vertical velocity to its horizontal velocity:
$$ 	an\phi = \frac{v_y}{v_x} $$

Now, let's consider a pebble thrown at an arbitrary angle $\alpha$ with the horizontal. Because there is no air resistance, its horizontal velocity remains perfectly constant throughout the flight:
$$ v_x = u \cos\alpha $$

What about the vertical velocity? We can use the third equation of motion. The initial vertical velocity is $u \sin\alpha$, and it falls through a vertical displacement $h$ under gravity $g$.

So, the vertical velocity $v_y$ at the ground is given by:
$$ v_y = \sqrt{(u \sin\alpha)^2 + 2gh} $$

Substituting these into our tangent equation, we get:
$$ 	an\phi = \frac{\sqrt{u^2 \sin^2\alpha + 2gh}}{u \cos\alpha} $$

### Minimizing the Impact Angle

To make this equation easier to handle, let's square both sides. This removes the pesky square root:
$$ 	an^2\phi = \frac{u^2 \sin^2\alpha + 2gh}{u^2 \cos^2\alpha} $$

Now, let's perform some algebraic magic. We can split the fraction into two parts:
$$ 	an^2\phi = \frac{u^2 \sin^2\alpha}{u^2 \cos^2\alpha} + \frac{2gh}{u^2 \cos^2\alpha} $$

The first term simplifies beautifully to $	an^2\alpha$. For the second term, remember that $\frac{1}{\cos^2\alpha}$ is $\sec^2\alpha$, which is identical to $1 + 	an^2\alpha$.

Substituting this in, we get:
$$ 	an^2\phi = 	an^2\alpha + \frac{2gh}{u^2}(1 + 	an^2\alpha) $$

Let's group the $	an^2\alpha$ terms together to see the structure clearly:
$$ 	an^2\phi = 	an^2\alpha \left(1 + \frac{2gh}{u^2}ight) + \frac{2gh}{u^2} $$

This is our master equation! We know that the minimum possible value for $\phi$ is $	heta$. To minimize $\phi$, we must minimize $	an^2\phi$.

Look at the right side of the equation. The terms $g$, $h$, and $u$ are constants. The only variable we control is the angle of projection, $\alpha$.

Since $	an^2\alpha$ can never be negative, its absolute minimum value is exactly $0$. This happens when $\alpha = 0^\circ$.

Physically, this means the shallowest impact angle occurs when the pebble is thrown **perfectly horizontally**!

### Final Calculation

Now that we know the minimum angle occurs when $\alpha = 0^\circ$, let's substitute this back into our equation.

The entire first term vanishes, and we replace $\phi$ with our minimum angle $	heta$:
$$ 	an^2	heta = \frac{2gh}{u^2} $$

The final step is just a simple rearrangement to isolate $h$. Multiply both sides by $u^2$ and divide by $2g$:
$$ h = \frac{u^2 	an^2	heta}{2g} $$

And there we have it! A beautiful, elegant expression for the height of the window, derived purely from the geometry of the trajectories.

Visualized Solution

\text{Setup}

\text{Angle of Impact}

\text{Velocity Components}

\text{Substituting Values}

\text{Simplifying the Expression}

\text{Rearranging Terms}

\text{Minimizing the Angle}

\text{Applying the Minimum Condition}

\text{Final Expression for Height}

\text{The Way Forward}

The Sigma Insight: Projectile Motion

Analyzing the Setup

The Master Equation

Minimizing the Impact Angle

Final Calculation

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