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 break down the motion of the ball and analyze the time taken for it to reach the floor after colliding with the ceiling of the cabin.

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

- The vertical component \( u_y \) is given by:
\[ u_y = u \sin \theta = 20 \sin(37^\circ) \]
- The horizontal component \( u_x \) is given by:
\[ u_x = u \cos \theta = 20 \cos(37^\circ) \]

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

### Step 2: Calculate the time to reach the ceiling
The height of the cabin is \( h = 7 \, \text{m} \). We need to find the time \( t_1 \) it takes for the ball to reach the ceiling using the following kinematic equation:
\[ s = u_y t + \frac{1}{2} a t^2 \]
Here, \( s = 7 \, \text{m} \), \( u_y = 12 \, \text{m/s} \), and \( a = -g = -10 \, \text{m/s}^2 \) (acceleration due to gravity).

Substituting the values:
\[ 7 = 12 t - 5 t^2 \]
Rearranging gives:
\[ 5t^2 - 12t + 7 = 0 \]

### Step 3: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 5 \cdot 7}}{2 \cdot 5} \]
Calculating the discriminant:
\[ = \sqrt{144 - 140} = \sqrt{4} = 2 \]
Thus, the time values are:
\[ t = \frac{12 \pm 2}{10} \]
Calculating both possible times:
1. \( t_1 = \frac{14}{10} = 1.4 \, \text{s} \)
2. \( t_2 = \frac{10}{10} = 1.0 \, \text{s} \)

### Step 4: Determine the time taken after collision with the ceiling
After reaching the ceiling, the ball will reverse its vertical velocity and fall back down. The velocity just before hitting the ceiling is \( v_y = u_y - gt_1 \):
\[ v_y = 12 - 10 \cdot 1.4 = -1 \, \text{m/s} \]
Now, we need to find the time \( t_3 \) it takes to fall back to the floor from the ceiling. The ball will fall a distance of \( 7 \, \text{m} \) with an initial velocity of \( 1 \, \text{m/s} \) (downwards).

Using the same kinematic equation:
\[ s = v_y t + \frac{1}{2} g t^2 \]
Substituting \( s = 7 \, \text{m} \), \( v_y = -1 \, \text{m/s} \), and \( g = 10 \, \text{m/s}^2 \):
\[ 7 = -1 t + 5 t^2 \]
Rearranging gives:
\[ 5t^2 + t - 7 = 0 \]

### Step 5: Solve the new quadratic equation
Using the quadratic formula again:
\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 5 \cdot (-7)}}{2 \cdot 5} \]
Calculating the discriminant:
\[ = \sqrt{1 + 140} = \sqrt{141} \]
Thus, the time values are:
\[ t = \frac{-1 \pm \sqrt{141}}{10} \]
Calculating the positive root gives:
\[ t \approx \frac{-1 + 11.83}{10} \approx 0.983 \, \text{s} \]

### Final Step: Total time after collision with the ceiling
The total time taken by the ball to reach the floor after colliding with the ceiling is:
\[ t_{\text{total}} = t_1 + t_3 \approx 1.4 + 0.983 \approx 2.383 \, \text{s} \]

### Final Answer
The time taken by the ball to reach the floor after the collision with the ceiling is approximately \( 2.38 \, \text{s} \).