A stone is dropped from a height h. simultaneously, another stone is thrown up from the ground which reaches a height 4h. The two stones will cross each other after time:-

Options

A. `sqrt((h)/(8g))`
B. `sqrt(8g)`
C. `sqrt(2gh)`
D. `sqrt((h)/(2g))`

Answer

To solve the problem of when two stones cross each other, we can break it down step by step.

### Step 1: Understand the Problem
We have two stones:
- Stone A is dropped from a height \( h \).
- Stone B is thrown upwards from the ground and reaches a maximum height of \( 4h \).

### Step 2: Determine the Initial Velocity of Stone B
Stone B reaches a height of \( 4h \). Using the kinematic equation:
\[ v^2 = u^2 - 2g s \]
where:
- \( v = 0 \) (velocity at maximum height),
- \( u \) is the initial velocity of Stone B,
- \( g \) is the acceleration due to gravity,
- \( s = 4h \) (the distance traveled upwards).

Substituting the values into the equation:
\[ 0 = u^2 - 2g(4h) \]
This simplifies to:
\[ u^2 = 8gh \]
Thus, the initial velocity \( u \) of Stone B is:
\[ u = \sqrt{8gh} \]

### Step 3: Write the Equations of Motion for Both Stones
1. For Stone A (dropped from height \( h \)):
- The distance fallen after time \( t \) is given by:
\[ H_1 = \frac{1}{2} g t^2 \]
(This distance is downward, so we consider it positive.)

2. For Stone B (thrown upwards):
- The distance traveled upwards after time \( t \) is:
\[ H_2 = ut - \frac{1}{2} g t^2 \]
Substituting \( u = \sqrt{8gh} \):
\[ H_2 = \sqrt{8gh} t - \frac{1}{2} g t^2 \]

### Step 4: Set Up the Equation for When They Cross
At the moment they cross each other, the total distance covered by both stones should equal the height \( h \):
\[ H_1 + H_2 = h \]
Substituting the expressions for \( H_1 \) and \( H_2 \):
\[ \frac{1}{2} g t^2 + \left(\sqrt{8gh} t - \frac{1}{2} g t^2\right) = h \]
This simplifies to:
\[ \sqrt{8gh} t = h \]

### Step 5: Solve for Time \( t \)
Rearranging gives:
\[ t = \frac{h}{\sqrt{8gh}} = \frac{h}{\sqrt{8g} \sqrt{h}} = \frac{\sqrt{h}}{\sqrt{8g}} = \sqrt{\frac{h}{8g}} \]

### Final Result
The time at which the two stones cross each other is:
\[ t = \sqrt{\frac{h}{8g}} \]

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