A ball is projected from a point in a horizontal plane with a velocity of 20 m/s at `30^(@)` with the horizontal. Coefficient of restitution is 0.8, then

Options

A. duration for which ball will rebound is 20 s
B. duration for which ball will rebound is 10 s
C. horizontal component of velocity of the ball is 17.3 m/s.
D. horizontal distance before the ball ceases to rebound is 173 m

Answer

To solve the problem step by step, we will analyze the motion of the ball projected at an angle and its behavior upon collision with the ground.

### Step 1: Resolve the Initial Velocity
The initial velocity \( u \) of the ball is given as 20 m/s at an angle of \( 30^\circ \) with the horizontal. We need to resolve this velocity into its horizontal and vertical components.

- **Horizontal Component**:
\[ u_x = u \cdot \cos(30^\circ) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

- **Vertical Component**:
\[ u_y = u \cdot \sin(30^\circ) = 20 \cdot \frac{1}{2} = 10 \, \text{m/s} \]

### Step 2: Calculate the Time of Flight Before First Impact
The time of flight \( T_0 \) for the first projectile motion can be calculated using the formula:
\[ T_0 = \frac{2u_y}{g} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

Substituting the values:
\[ T_0 = \frac{2 \cdot 10}{9.81} \approx 2.04 \, \text{s} \]

### Step 3: Analyze the Collision
Upon collision with the ground, only the vertical component of the velocity will change due to the coefficient of restitution \( e \).

- **Velocity just before impact**: The vertical component just before impact can be calculated using the formula:
\[ v_y = u_y - g \cdot T_0 \]
Substituting the values:
\[ v_y = 10 - 9.81 \cdot 2.04 \approx -9.81 \, \text{m/s} \]

- **Velocity after impact**: The vertical component after the impact will be:
\[ v'_y = -e \cdot v_y = -0.8 \cdot (-9.81) \approx 7.848 \, \text{m/s} \]

### Step 4: Calculate the Total Time of Flight
After the first bounce, the time of flight for the subsequent bounces will be reduced by the coefficient of restitution. The time of flight for the second bounce can be calculated as:
\[ T_1 = \frac{2v'_y}{g} = \frac{2 \cdot 7.848}{9.81} \approx 1.60 \, \text{s} \]

The time of flight for each subsequent bounce will continue to decrease by a factor of \( e \). Thus, the total time of flight \( T \) can be calculated as an infinite geometric series:
\[ T = T_0 + eT_0 + e^2T_0 + \ldots \]
This series can be summed up using the formula for the sum of an infinite geometric series:
\[ S = \frac{a}{1 - r} \]
where \( a = T_0 \) and \( r = e \).

Substituting the values:
\[ T = \frac{T_0}{1 - e} = \frac{2.04}{1 - 0.8} = \frac{2.04}{0.2} = 10.2 \, \text{s} \]

### Step 5: Calculate the Horizontal Distance
The horizontal distance \( D \) traveled can be calculated using the horizontal component of the velocity and the total time of flight:
\[ D = u_x \cdot T = 10\sqrt{3} \cdot 10.2 \approx 17.32 \cdot 10.2 \approx 176.8 \, \text{m} \]

### Summary of Results
- The total time of flight is approximately \( 10.2 \, \text{s} \).
- The horizontal distance traveled is approximately \( 176.8 \, \text{m} \).