A ball is projected from the floor of a cabin of height 7m with speed 20 m/s at an angle of `37^(@)` with the floor of cabin. It makes a successive collision with the wall of cabin and then return again to its floor. Assume all collisions are perfectly elastic. Find time taken (in sec) by the ball to reach the floor after collisions with the ceiling of cabin.

Answer

To solve the problem step-by-step, we will analyze the motion of the ball after it is projected, calculate the time taken to reach the ceiling, and then determine the time taken to return to the floor after the collision with the ceiling.

### Step 1: Break down the initial velocity into components
The ball is projected with a speed \( u = 20 \, \text{m/s} \) at an angle \( \theta = 37^\circ \) with the horizontal. We can find the horizontal and vertical components of the initial velocity:

- Horizontal component \( u_x = u \cos(\theta) = 20 \cos(37^\circ) \)
- Vertical component \( u_y = u \sin(\theta) = 20 \sin(37^\circ) \)

Using \( \cos(37^\circ) \approx 0.8 \) and \( \sin(37^\circ) \approx 0.6 \):
- \( u_x = 20 \times 0.8 = 16 \, \text{m/s} \)
- \( u_y = 20 \times 0.6 = 12 \, \text{m/s} \)

### Step 2: Calculate the time taken to reach the ceiling
The height of the cabin is \( h = 7 \, \text{m} \). We can use the kinematic equation for vertical motion to find the time \( t_1 \) taken to reach the ceiling:

\[ h = u_y t_1 - \frac{1}{2} g t_1^2 \]

Where \( g = 10 \, \text{m/s}^2 \). Plugging in the values:

\[ 7 = 12 t_1 - \frac{1}{2} \cdot 10 \cdot t_1^2 \]

This simplifies to:

\[ 7 = 12 t_1 - 5 t_1^2 \]

Rearranging gives:

\[ 5 t_1^2 - 12 t_1 + 7 = 0 \]

### Step 3: Solve the quadratic equation
Now we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 5, b = -12, c = 7 \):

\[ t_1 = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 5 \cdot 7}}{2 \cdot 5} \]
\[ t_1 = \frac{12 \pm \sqrt{144 - 140}}{10} \]
\[ t_1 = \frac{12 \pm \sqrt{4}}{10} \]
\[ t_1 = \frac{12 \pm 2}{10} \]

Calculating the two possible values:

1. \( t_1 = \frac{14}{10} = 1.4 \, \text{s} \)
2. \( t_1 = \frac{10}{10} = 1.0 \, \text{s} \)

Since the ball will reach the ceiling first, we take \( t_1 = 1.4 \, \text{s} \).

### Step 4: Calculate the velocity just before the collision with the ceiling
Using the vertical motion equation:

\[ v_y = u_y - g t_1 \]
\[ v_y = 12 - 10 \cdot 1.4 = 12 - 14 = -2 \, \text{m/s} \]

The negative sign indicates that the ball is moving downward just before the collision.

### Step 5: Determine the velocity after the elastic collision
In a perfectly elastic collision, the ball will rebound with the same speed but in the opposite direction. Thus, the velocity after the collision is:

\[ v_{y, \text{after}} = 2 \, \text{m/s} \quad \text{(upward)} \]

### Step 6: Calculate the time taken to reach the floor after the collision
Now we need to find the time \( t_2 \) it takes to fall from the ceiling back to the floor. Using the kinematic equation again:

\[ s = v_{y, \text{after}} t_2 + \frac{1}{2} g t_2^2 \]
\[ 7 = 2 t_2 + \frac{1}{2} \cdot 10 t_2^2 \]
\[ 7 = 2 t_2 + 5 t_2^2 \]

Rearranging gives:

\[ 5 t_2^2 + 2 t_2 - 7 = 0 \]

### Step 7: Solve the quadratic equation for \( t_2 \)
Using the quadratic formula again:

\[ t_2 = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 5 \cdot (-7)}}{2 \cdot 5} \]
\[ t_2 = \frac{-2 \pm \sqrt{4 + 140}}{10} \]
\[ t_2 = \frac{-2 \pm \sqrt{144}}{10} \]
\[ t_2 = \frac{-2 \pm 12}{10} \]

Calculating the two possible values:

1. \( t_2 = \frac{10}{10} = 1.0 \, \text{s} \)
2. \( t_2 = \frac{-14}{10} \) (not possible since time cannot be negative)

Thus, the time taken by the ball to reach the floor after the collision with the ceiling is:

\[ t_2 = 1.0 \, \text{s} \]

### Final Answer
The total time taken by the ball to reach the floor after the collision with the ceiling is \( 1.0 \, \text{s} \).

---