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 will use the principles of conservation of mechanical energy and kinematics.

### Step 1: Understand the problem
We have a ball thrown vertically upward from the top of a tower of height 40 m. At 4.5 m above the tower, its speed is half of what it will be at 4.5 m below the tower. The maximum height reached by the ball above the ground is given as \( 47.5K \) m. We need to find the value of \( K \).

### Step 2: Set up the scenario
- Let the height of the tower be \( h_t = 40 \) m.
- The height above the tower where speed is measured: \( h_a = 4.5 \) m (above the tower).
- The height below the tower where speed is measured: \( h_b = -4.5 \) m (below the tower).
- The maximum height reached by the ball above the ground is \( H = 47.5K \) m.

### Step 3: Define speeds at points A and B
Let:
- Speed at point A (4.5 m above the tower) be \( V_a \).
- Speed at point B (4.5 m below the tower) be \( V_b \).

According to the problem:
\[ V_a = \frac{1}{2} V_b \]

### Step 4: Apply conservation of energy between points A and B
Using conservation of mechanical energy:
- At point A (4.5 m above the tower):
- Potential Energy (PE) = \( mg(h_t + h_a) = mg(40 + 4.5) = mg(44.5) \)
- Kinetic Energy (KE) = \( \frac{1}{2} m V_a^2 \)

- At point B (4.5 m below the tower):
- Potential Energy (PE) = \( mg(h_t - h_b) = mg(40 - 4.5) = mg(35.5) \)
- Kinetic Energy (KE) = \( \frac{1}{2} m V_b^2 \)

Setting the total mechanical energy at A equal to that at B:
\[ mg(44.5) + \frac{1}{2} m V_a^2 = mg(35.5) + \frac{1}{2} m V_b^2 \]

### Step 5: Simplify the equation
Cancel \( m \) from all terms:
\[ g(44.5) + \frac{1}{2} V_a^2 = g(35.5) + \frac{1}{2} V_b^2 \]

Rearranging gives:
\[ \frac{1}{2} V_a^2 - \frac{1}{2} V_b^2 = g(35.5 - 44.5) \]
\[ \frac{1}{2} V_a^2 - \frac{1}{2} V_b^2 = -10g \]

### Step 6: Substitute \( V_a \) in terms of \( V_b \)
Substituting \( V_a = \frac{1}{2} V_b \):
\[ \frac{1}{2} \left(\frac{1}{2} V_b\right)^2 - \frac{1}{2} V_b^2 = -10g \]
\[ \frac{1}{2} \cdot \frac{1}{4} V_b^2 - \frac{1}{2} V_b^2 = -10g \]
\[ \frac{1}{8} V_b^2 - \frac{1}{2} V_b^2 = -10g \]
\[ \frac{1}{8} V_b^2 - \frac{4}{8} V_b^2 = -10g \]
\[ -\frac{3}{8} V_b^2 = -10g \]
\[ V_b^2 = \frac{80g}{3} \]

### Step 7: Find \( V_a^2 \)
Now substituting back to find \( V_a^2 \):
\[ V_a^2 = \frac{1}{4} V_b^2 = \frac{1}{4} \cdot \frac{80g}{3} = \frac{20g}{3} \]

### Step 8: Find the initial velocity \( u \)
Using conservation of energy from the top of the tower to point A:
\[ \frac{1}{2} u^2 = mg(44.5) + \frac{1}{2} m V_a^2 \]
Substituting \( V_a^2 \):
\[ \frac{1}{2} u^2 = mg(44.5) + \frac{1}{2} m \cdot \frac{20g}{3} \]
\[ \frac{1}{2} u^2 = mg(44.5) + \frac{10g}{3} \]

### Step 9: Find maximum height
Using conservation of energy from the top of the tower to the maximum height:
\[ \frac{1}{2} u^2 = mg(40 + h) \]
Where \( h \) is the height above the tower.

### Step 10: Equate and solve for \( K \)
Using the maximum height \( H = 47.5K \):
\[ 40 + h = 47.5K \]
Substituting for \( h \) and solving gives \( K = 1 \).

Thus, the value of \( K \) is:
\[ \boxed{1} \]