A ball is thrown upwards with a speed of `40m//s`. When the speed becomes half of the initial speed, gravity is switched off for next 2 second. After that gravity is again switched on but magnitude gravity is doubled. The total distance travelled by the ball from `t=0` to the time when the ball reaches the maximum heighth is `55beta`. Find the value of `beta`.

Answer

To solve the problem step by step, we will analyze the motion of the ball in three phases:

1. **Phase 1: Ball thrown upwards until speed is halved**
2. **Phase 2: Gravity is switched off for 2 seconds**
3. **Phase 3: Gravity is switched on with doubled magnitude until maximum height is reached**

### Step 1: Calculate the distance traveled until the speed is halved

The ball is thrown upwards with an initial speed \( u = 40 \, \text{m/s} \). We need to find the distance traveled until the speed becomes half of the initial speed, which is \( v = 20 \, \text{m/s} \).

Using the equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 20 \, \text{m/s} \)
- \( u = 40 \, \text{m/s} \)
- \( a = -g = -10 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ (20)^2 = (40)^2 + 2(-10)s \]
\[ 400 = 1600 - 20s \]
Rearranging gives:
\[ 20s = 1600 - 400 = 1200 \]
\[ s = \frac{1200}{20} = 60 \, \text{m} \]

### Step 2: Calculate the distance traveled during the 2 seconds of no gravity

After reaching a speed of \( 20 \, \text{m/s} \), gravity is turned off for 2 seconds. The distance traveled during this time with constant speed is:
\[ s' = v \cdot t = 20 \, \text{m/s} \cdot 2 \, \text{s} = 40 \, \text{m} \]

### Step 3: Calculate the distance traveled until maximum height with doubled gravity

After 2 seconds, gravity is switched back on, but its magnitude is doubled, so \( a = -2g = -20 \, \text{m/s}^2 \). The initial speed at this point is still \( 20 \, \text{m/s} \). We need to find the distance traveled until the ball reaches maximum height (where final speed \( v = 0 \)).

Using the same equation of motion:
\[ v^2 = u^2 + 2as \]
Substituting the values:
\[ 0 = (20)^2 + 2(-20)s \]
\[ 0 = 400 - 40s \]
Rearranging gives:
\[ 40s = 400 \]
\[ s = \frac{400}{40} = 10 \, \text{m} \]

### Step 4: Calculate the total distance traveled

Now, we can sum up all the distances:
\[ \text{Total distance} = s + s' + s'' = 60 \, \text{m} + 40 \, \text{m} + 10 \, \text{m} = 110 \, \text{m} \]

### Step 5: Relate total distance to the given expression

According to the problem, the total distance is given as \( 55\beta \):
\[ 55\beta = 110 \]
Solving for \( \beta \):
\[ \beta = \frac{110}{55} = 2 \]

### Final Answer
The value of \( \beta \) is \( \boxed{2} \).

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