A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle o

Q: Why do we completely ignore the 10 m height of the building in our calculation?

Because the question asks for the horizontal distance when the ball is at a height of $10\text{ m}$ from the ground. Since it was thrown from a height of $10\text{ m}$, the vertical displacement between the start and the target point is zero. The ball behaves exactly as if it were thrown on a flat ground until it passes that point.

Q: What if the question asked for the horizontal distance when the ball hits the ground?

In that case, the vertical displacement would be $-10\text{ m}$. You could not use the simple range formula. You would have to use the second equation of motion ($S_y = u_y t + \frac{1}{2} a_y t^2$) to find the total time of flight, and then multiply that time by the horizontal velocity ($u_x$) to find the total horizontal distance.

Q: Why is the angle $2\theta$ in the range formula?

The range formula is derived by multiplying the horizontal velocity ($u \cos\theta$) by the time of flight ($\frac{2u \sin\theta}{g}$). This gives $\frac{u^2 (2 \sin\theta \cos\theta)}{g}$. Using the trigonometric identity $2 \sin\theta \cos\theta = \sin 2\theta$, we get the final formula $\frac{u^2 \sin 2\theta}{g}$.

Q: Can we solve this using the basic equations of motion instead of the direct formula?

Absolutely! You can find the time it takes to return to the same height by setting vertical displacement $S_y = 0$ in $S_y = (u \sin\theta)t - \frac{1}{2}gt^2$. Solving for $t$ gives $t = \frac{2u \sin\theta}{g}$. Then, multiply this time by the horizontal velocity $u_x = u \cos\theta$ to get the distance. It will yield the exact same result.

Q: What is the significance of $\sqrt{3} \approx 1.732$?

It's a standard mathematical constant that frequently appears in geometry and physics problems involving $30^\circ$ and $60^\circ$ angles. Memorizing the approximate values of $\sqrt{2} \approx 1.414$ and $\sqrt{3} \approx 1.732$ is highly recommended for competitive exams to quickly evaluate final numerical answers.

Question

A boy playing on the roof of a $10	ext{ m}$ high building throws a ball with a speed of $10	ext{ m/s}$ at an angle of $30^\circ$ with the horizontal. How far from the throwing point will the ball be at the height of $10	ext{ m}$ from the ground?
$[g = 10	ext{ m/s}^2, \sin 30^\circ = \frac{1}{2}, \cos 30^\circ = \frac{\sqrt{3}}{2}]$

Options:

(A) $5.20	ext{ m}$

(B) $4.33	ext{ m}$

(C) $2.60	ext{ m}$

(D) $8.66	ext{ m}$

Sigma Percentile · Accepted Answer

### The Illusion of Height: Solving Projectile Motion on a Building

Imagine you are standing on the roof of a $10	ext{ m}$ high building. You have a ball in your hand, and you throw it with a speed of $10	ext{ m/s}$ at an angle of $30^\circ$ above the horizontal. The ball soars into the air, traces a beautiful parabolic arc, and eventually plummets to the ground below.

But the question asks us something very specific: How far from the throwing point will the ball be when it is exactly at a height of $10	ext{ m}$ from the ground?

At first glance, the $10	ext{ m}$ height of the building seems like a crucial piece of information. However, physics often rewards us for looking at problems from the right perspective.

### Analyzing the Setup

Let's define our coordinate system. Let the point of projection on the roof be our origin, $O(0,0)$. The ball is thrown from this point, goes up, reaches a maximum height, and then comes back down.

We are asked to find the horizontal distance when the ball is at a height of $10	ext{ m}$ from the ground. Since the building itself is $10	ext{ m}$ high, the point of projection $O$ is also at a height of $10	ext{ m}$ from the ground.

This means we are looking for a point $P$ on the trajectory that has the exact same vertical elevation as our starting point $O$. The vertical displacement between $O$ and $P$ is exactly zero!

### The Master Equation

Because the initial and final heights are identical, the segment of the trajectory from $O$ to $P$ is mathematically indistinguishable from a projectile launched on a perfectly flat, level plane. The $10	ext{ m}$ drop to the ground below doesn't affect the ball's motion until *after* it passes point $P$.

Therefore, the horizontal distance $OP$ is simply the standard horizontal range $R$ of a projectile. We can completely ignore the height of the building for this specific calculation.

The formula for the horizontal range of a projectile on a level plane is:

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

Where:
- $u$ is the initial velocity ($10	ext{ m/s}$)
- $	heta$ is the angle of projection ($30^\circ$)
- $g$ is the acceleration due to gravity ($10	ext{ m/s}^2$)

### Final Calculation

Now, it's just a matter of substituting our known values into the master equation. Let's plug them in:

$$R = \frac{(10)^2 \sin(2 	imes 30^\circ)}{10}$$

First, let's simplify the velocity and gravity terms. $10$ squared is $100$, and dividing by $10$ gives us $10$:

$$R = 10 	imes \sin 60^\circ$$

Next, we evaluate the trigonometric function. We know from standard trigonometric values that $\sin 60^\circ = \frac{\sqrt{3}}{2}$. Substituting this back in:

$$R = 10 	imes \frac{\sqrt{3}}{2}$$

$$R = 5\sqrt{3}$$

To find the numerical value, we use the approximation $\sqrt{3} \approx 1.732$:

$$R = 5 	imes 1.732 = 8.66	ext{ m}$$

**The ball will be $8.66	ext{ m}$ away horizontally when it returns to the $10	ext{ m}$ height.**

This problem is a classic example of how identifying symmetries in physics—in this case, the symmetry of returning to the initial height—can drastically simplify the math. Always look for the hidden simplicity before diving into complex equations!

Select Answer:

Visualized Solution

Visualizing the Trajectory

The Level Plane Equivalence

Setting up the Range Formula

Evaluating the Velocity Term

Evaluating the Trigonometric Term

Final Calculation

The Way Forward

The Sigma Insight: Projectile Motion

The Illusion of Height

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

A boy can throw a stone upto a maximum height of $10 m$ . The maximum horizontal distance that the boy can throw the same stone upto will be

A player kicks a football with an initial speed of $25 ms^{- 1}$ at an angle of $4 5^{\circ}$ from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion ? (Take, $g = 10 ms^{- 2}$ )

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Select Answer:

Visualized Solution

Visualizing the Trajectory

The Level Plane Equivalence

Setting up the Range Formula

Evaluating the Velocity Term

Evaluating the Trigonometric Term

Final Calculation

The Way Forward

The Sigma Insight: Projectile Motion

The Illusion of Height

Analyzing the Setup

The Master Equation

Final Calculation

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