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

Options

A. `vecv//2"and"v//2`
B. 0 and v / 2
C. 0 and 0
D. `vecv//2"and"0`

Answer

To solve the problem where a stone is dropped from a height \( h \) and another stone is thrown upwards from the ground to reach a height of \( 4h \), we will follow these steps:

### Step 1: Determine the initial velocity of the second stone
The second stone reaches a maximum height of \( 4h \). At the maximum height, its final velocity is \( 0 \). We can use the conservation of energy principle to find the initial velocity \( u \) of the second stone.

Using the equation for potential energy and kinetic energy:
\[ \text{Potential Energy} = \text{Kinetic Energy} \]
\[ mgh = \frac{1}{2} mv^2 \]
Substituting \( h = 4h \):
\[ mg(4h) = \frac{1}{2} m u^2 \]
Canceling \( m \) from both sides:
\[ 4gh = \frac{1}{2} u^2 \]
Multiplying both sides by \( 2 \):
\[ 8gh = u^2 \]
Taking the square root:
\[ u = \sqrt{8gh} \]

### Step 2: Write the equations of motion for both stones
For the first stone (dropped from height \( h \)):
- Initial velocity \( u_1 = 0 \)
- Displacement after time \( t \):
\[ y_1 = h - \frac{1}{2} g t^2 \]

For the second stone (thrown upwards):
- Initial velocity \( u_2 = \sqrt{8gh} \)
- Displacement after time \( t \):
\[ y_2 = u_2 t - \frac{1}{2} g t^2 \]
Substituting \( u_2 \):
\[ y_2 = \sqrt{8gh} t - \frac{1}{2} g t^2 \]

### Step 3: Set the heights equal at the crossing point
At the point where both stones cross each other, their heights will be equal:
\[ y_1 = y_2 \]
Substituting the equations:
\[ h - \frac{1}{2} g t^2 = \sqrt{8gh} t - \frac{1}{2} g t^2 \]
Canceling \( -\frac{1}{2} g t^2 \) from both sides:
\[ h = \sqrt{8gh} t \]

### Step 4: Solve for time \( t \)
Rearranging the equation:
\[ t = \frac{h}{\sqrt{8gh}} \]
This simplifies to:
\[ t = \frac{1}{\sqrt{8g}} \sqrt{h} \]

### Summary
The time at which the two stones cross each other is given by:
\[ t = \frac{1}{\sqrt{8g}} \sqrt{h} \]