A ball is thrown vertically up with a certain velocity from the top of a tower of height 40m. At 4.5m above the top of the tower its speed is exactly half of that it will have at 4.5m below the top of the tower. The maximum height reached by the ball aboe the ground is 47.5K m. Find K.

Answer

To solve the problem step by step, we need to analyze the motion of the ball thrown from the top of the tower and apply the kinematic equations appropriately.

### Step 1: Understand the Problem
The ball is thrown upwards from the top of a tower of height 40 m. We need to find the maximum height reached by the ball above the ground, which is given as \( 47.5K \) m. We also know that at a height of \( 4.5 \) m above the tower, the speed of the ball is half of what it will be \( 4.5 \) m below the top of the tower.

### Step 2: Define Variables
- Let the initial velocity of the ball be \( u \).
- The height of the tower is \( h_t = 40 \) m.
- The height above the tower where the speed is measured is \( h_1 = 4.5 \) m.
- The height below the tower where the speed is measured is \( h_2 = -4.5 \) m (considering downward direction as negative).

### Step 3: Apply Kinematic Equations
1. **At height \( 4.5 \) m above the tower**:
- Using the third equation of motion:
\[ v_1^2 = u^2 - 2g h_1 \]
where \( v_1 \) is the velocity at \( 4.5 \) m above the tower.

2. **At height \( 4.5 \) m below the tower**:
- Using the third equation of motion:
\[ v_2^2 = u^2 - 2g (-h_2) = u^2 + 2g h_2 \]
where \( v_2 \) is the velocity at \( 4.5 \) m below the tower.

### Step 4: Relate the Velocities
From the problem statement, we know:
\[ v_1 = \frac{1}{2} v_2 \]
Substituting \( v_1 \) and \( v_2 \) into this equation gives:
\[ u^2 - 2g h_1 = \frac{1}{4}(u^2 + 2g h_2) \]

### Step 5: Substitute Known Values
Substituting \( h_1 = 4.5 \) m and \( h_2 = 4.5 \) m into the equation:
\[ u^2 - 2g(4.5) = \frac{1}{4}(u^2 + 2g(4.5)) \]

### Step 6: Solve for \( u^2 \)
Multiply through by 4 to eliminate the fraction:
\[ 4u^2 - 8g(4.5) = u^2 + 2g(4.5) \]
Rearranging gives:
\[ 3u^2 = 10g(4.5) \]
Thus,
\[ u^2 = \frac{10g(4.5)}{3} \]

### Step 7: Find Maximum Height
The maximum height \( h_{max} \) above the ground is given by:
\[ h_{max} = h_t + h_1 + h \]
where \( h \) is the additional height gained after reaching the maximum point.

Using the first equation of motion to find \( h \):
\[ 0 = u^2 - 2gh \implies h = \frac{u^2}{2g} \]
Substituting \( u^2 \):
\[ h = \frac{10g(4.5)}{3 \cdot 2g} = \frac{10 \cdot 4.5}{6} = 7.5 \text{ m} \]

### Step 8: Calculate Total Maximum Height
Now, substituting back to find \( h_{max} \):
\[ h_{max} = 40 + 4.5 + 7.5 = 52 \text{ m} \]
Given \( h_{max} = 47.5K \), we set:
\[ 52 = 47.5K \implies K = \frac{52}{47.5} = 1.0947 \approx 1 \]

### Final Answer
Thus, the value of \( K \) is approximately \( 1 \).