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Step-by-Step Solution
Step 1: Understand the Problem
A stone (stone 1) is dropped from the top of a building. At the instant it crosses a point 5 m below the top, another stone (stone 2) starts to fall from a point 25 m below the top. Both stones then reach the bottom of the building at the same time. We want to determine the total height of the building.
Step 2: Define Variables
Let the total height of the building be $H$.
The second stone is released 25 m below the top, so the remaining distance for stone 2 to travel from its starting point to the bottom is $h$. Later, we will find that $h = 20\,\text{m}$, and hence the total height $H = 25 + h.$
Let $t$ be the time (after stone 2 is released) it takes for both stones to travel their respective remaining distances to the bottom and arrive simultaneously.
Assume the acceleration due to gravity $g = 10\,\text{m/s}^2$ (as used in the provided solution).
Step 3: Set Up the Equations
According to the provided reference solution, the equations are set as follows:
For stone 1 (particle 1):
When stone 1 is at 5 m below the top, it has already traveled those 5 m. From that point downward, it still has $(H - 5)$ m to go. However, in the provided reference, the distance is broken into $20 + h$, matching how the solution is labeling segments. The reference solution writes:
$20 + h = 10t + \frac{1}{2} g t^2$
(i)
Where $10t$ is the initial velocity $v$ times $t$ (based on the moment it crosses the 5 m point) plus the distance covered under constant acceleration for $t$ seconds. In the simplification used, $20 + h$ appears as the distance yet to be covered by stone 1 from that crossing point.
For stone 2 (particle 2):
The second stone starts from rest at 25 m below the top, and from that point, it has a distance $h$ to travel. Its motion is given by the standard free-fall formula:
$h = \frac{1}{2} g t^2$
(ii)
Step 4: Solve for Time
We substitute equation (ii) into equation (i) to eliminate $h$:
$20 + \frac{1}{2} g t^2 = 10t + \frac{1}{2} g t^2$
Canceling the $\frac{1}{2} g t^2$ terms on both sides gives:
$20 = 10t$
Therefore,
$t = 2 \,\text{s}.$
Step 5: Find the Distance $h$ for Stone 2
Using (ii):
$h = \frac{1}{2} g t^2 = \frac{1}{2} \times 10 \times (2)^2 = 20 \,\text{m}.$
Step 6: Calculate the Total Height of the Building
Since stone 2 was released 25 m below the top, that portion of the building is already known. We add 25 m to the distance $h = 20$ m that stone 2 traveled:
$H = 25 + 20 = 45\,\text{m}.$
Thus, the height of the building is 45 m.
Referenced Image from the Provided Solution
