From an elevated point P a stone is projected vertically upward. When it reaches a distance y below the point of projection its velocity is double the velocity when it was at a height y above p. The greatest height reached by it above P is

Options

A. `(2y)/(3)`
B. `(5y)/(3)`
C. `y/3`
D. 2y

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a stone projected vertically upward from point P. We need to find the maximum height (let's call it X) reached by the stone above point P. We know that when the stone is at a height Y below P, its velocity (let's call it V_B) is double the velocity (let's call it V_A) when it is at a height Y above P.

### Step 2: Set up the equations of motion
Using the equations of motion, we can express the velocities at the two points.

1. At the maximum height, the initial velocity (U) is 0, and the displacement (S) from the maximum height to the height Y above P is (X - Y). The equation of motion gives us:
\[ V_A^2 = 0 + 2g(X - Y) \quad \text{(1)} \]

2. For the point Y below P, the displacement from the maximum height to this point is (X + Y). The equation of motion gives us:
\[ V_B^2 = 0 + 2g(X + Y) \quad \text{(2)} \]

### Step 3: Relate the velocities
According to the problem, we have:
\[ V_B = 2V_A \]
Squaring both sides gives:
\[ V_B^2 = 4V_A^2 \]

### Step 4: Substitute the velocities into the equations
Substituting the expressions for \(V_A^2\) and \(V_B^2\) into the equation \(V_B^2 = 4V_A^2\):
\[ 2g(X + Y) = 4(2g(X - Y)) \]

### Step 5: Simplify the equation
Now we can simplify the equation:
\[ 2g(X + Y) = 8g(X - Y) \]
Dividing both sides by \(2g\) (assuming \(g \neq 0\)):
\[ X + Y = 4(X - Y) \]
Expanding the right side:
\[ X + Y = 4X - 4Y \]
Rearranging gives:
\[ X + Y + 4Y = 4X \]
\[ X + 5Y = 4X \]
\[ 5Y = 3X \]
So we find:
\[ X = \frac{5Y}{3} \]

### Step 6: Conclusion
The greatest height reached by the stone above point P is:
\[ \boxed{\frac{5Y}{3}} \]