A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another st

Q: Why can't we just equate their distances?

Because they start at different times and from different positions. The only constraint given in the problem that links their motions is that they hit the ground at the exact same time.

Q: What happens to the 'g' in the equation?

It cancels out! Because both stones are falling under the exact same gravitational acceleration, 'g' appears as a common factor in all the time expressions and can be divided out.

Q: Is there a faster way to solve this without the messy algebra?

Yes! Using relative motion. Once the second stone is dropped, their relative acceleration is zero. The first stone catches up to the second stone with a constant relative velocity. This makes the calculation much shorter.

Q: Why did we square the equation $\sqrt{x} - \sqrt{5} = \sqrt{x - 25}$?

To eliminate the square roots and solve for $x$. Squaring both sides is the standard algebraic technique for solving radical equations.

Q: What if the second stone was thrown downwards instead of dropped?

Then its initial velocity $u_2$ would not be zero. The time equation for the second stone would become a quadratic equation, making the algebra significantly more complex, though the core logic remains the same.

Question

A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25 m below the top. Both stones reach the bottom of building simultaneously. The height of the building is

Options:

(A) 45 m

(B) 25 m

(C) 35 m

(D) 50 m

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the top of a tall building of unknown height, let's call it $x$. You drop a red stone from the very top. Just as this red stone passes the $5	ext{ m}$ mark, a friend drops a blue stone from a window $25	ext{ m}$ below the top.

The problem states a crucial constraint: **both stones hit the ground simultaneously**. This means the time the red stone spends in the air *after* crossing the $5	ext{ m}$ mark must be exactly equal to the *total* time the blue stone spends in the air.

### The Master Equation

First, let's figure out the "head start" time of the red stone. Using the second equation of motion $s = ut + \frac{1}{2}gt^2$, the time taken to fall the first $5	ext{ m}$ is:

$$t_0 = \sqrt{\frac{2 	imes 5}{g}} = \sqrt{\frac{10}{g}}$$

The total time the red stone takes to fall the entire height $x$ is:

$$t_1 = \sqrt{\frac{2x}{g}}$$

Therefore, the remaining time for the red stone to reach the ground is $t_1 - t_0$.

Now, let's look at the blue stone. It is dropped from $25	ext{ m}$ below the top, so it only needs to fall a distance of $(x - 25)	ext{ m}$. Its total time of flight is:

$$t_2 = \sqrt{\frac{2(x - 25)}{g}}$$

Equating the remaining time of the red stone to the total time of the blue stone gives us our master equation:

$$\sqrt{\frac{2x}{g}} - \sqrt{\frac{10}{g}} = \sqrt{\frac{2(x - 25)}{g}}$$

### Final Calculation

Don't let the square roots and the $g$'s intimidate you. We can multiply the entire equation by $\sqrt{\frac{g}{2}}$ to beautifully cancel out the constants:

$$\sqrt{x} - \sqrt{5} = \sqrt{x - 25}$$

To solve for $x$, we square both sides. Be careful with the expansion on the left side:

$$x + 5 - 2\sqrt{5x} = x - 25$$

The $x$ terms cancel out perfectly. Rearranging the terms gives:

$$30 = 2\sqrt{5x} \implies 15 = \sqrt{5x}$$

Squaring both sides one last time yields:

$$225 = 5x \implies x = 45	ext{ m}$$

### The Elegant Alternative: Relative Motion

If you want to solve this like a pro, use relative kinematics! At the instant the second stone is dropped, the first stone is at $5	ext{ m}$ moving down with velocity $v_1 = \sqrt{10g}$. The second stone is at $25	ext{ m}$ with velocity $v_2 = 0$. The initial gap between them is $20	ext{ m}$.

Since both are in free fall, their relative acceleration is $a_{	ext{rel}} = g - g = 0$. The first stone approaches the second with a constant relative velocity of $v_{	ext{rel}} = \sqrt{10g}$. The time taken to close the $20	ext{ m}$ gap (which happens exactly at the ground) is $t = \frac{20}{\sqrt{10g}}$.

During this time, the second stone falls a distance $s_2 = \frac{1}{2}gt^2 = \frac{1}{2}g \left(\frac{400}{10g}ight) = 20	ext{ m}$. Since it started $25	ext{ m}$ below the top, the total height of the building is simply $25 + 20 = 45	ext{ m}$. Brilliant, isn't it?

Animated Solution for Physics - Kinematics: A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25 m below the top. Both stones reach the bottom of building simultaneously. The height of the building is

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Visualized Solution

\text{Visualizing the Setup}

\text{Time taken by Stone 1 to fall } 5\text{ m}

\text{Total Time for Stone 1}

\text{Time taken by Stone 2}

\text{Equating the Times}

\text{Simplifying the Equation}

\text{Solving for } x

\text{Final Calculation}

\text{Alternative Approach: Relative Motion}

The Sigma Insight: Motion in a Straight Line

Analyzing the Setup

The Master Equation

Final Calculation

The Elegant Alternative

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