A ball is thrown from the top of a tower in vertically upward direction. Velocity at a point h m below the point of projection is twice of the velocity at a point h m above the point of projection. Find the maximum height reached by the ball above the top of tower.

Options

A. ` 2h`
B. ` 3h`
C. ` (5//3) h`
D. ` (4//3)h`

Explanation

To solve the problem step by step, we will break down the information given and apply the relevant physics equations.

### Step 1: Understand the Problem
We have a ball thrown upwards from the top of a tower. The velocity at a point \( h \) meters below the point of projection is twice the velocity at a point \( h \) meters above the point of projection. We need to find the maximum height reached by the ball above the top of the tower.

### Step 2: Define Variables
- Let \( u \) be the initial velocity of the ball at point P (the top of the tower).
- Let \( v \) be the velocity at point A (h meters above P).
- The velocity at point B (h meters below P) is \( 2v \).

### Step 3: Apply Kinematic Equations
1. For point A (h meters above P):
Using the equation of motion:
\[ v^2 = u^2 - 2gh \]
(since the ball is moving upwards, we take g as negative).

2. For point B (h meters below P):
Using the equation of motion:
\[ (2v)^2 = u^2 - 2g(-h) \implies 4v^2 = u^2 + 2gh \]

### Step 4: Set Up the Equations
From the equations derived:
1. \( v^2 = u^2 - 2gh \) (Equation 1)
2. \( 4v^2 = u^2 + 2gh \) (Equation 2)

### Step 5: Substitute Equation 1 into Equation 2
Substituting \( v^2 \) from Equation 1 into Equation 2:
\[ 4(u^2 - 2gh) = u^2 + 2gh \]
Expanding this gives:
\[ 4u^2 - 8gh = u^2 + 2gh \]
Rearranging terms:
\[ 4u^2 - u^2 = 8gh + 2gh \]
\[ 3u^2 = 10gh \]
Thus,
\[ u^2 = \frac{10gh}{3} \quad (Equation 3) \]

### Step 6: Find Maximum Height
Using the formula for maximum height \( h_{max} \):
\[ h_{max} = \frac{u^2}{2g} \]
Substituting Equation 3 into this:
\[ h_{max} = \frac{\frac{10gh}{3}}{2g} = \frac{10h}{6} = \frac{5h}{3} \]

### Conclusion
The maximum height reached by the ball above the top of the tower is:
\[ \boxed{\frac{5h}{3}} \]