To solve the problem step by step, let's break down the given information and apply the relevant physics concepts.
### Step 1: Determine the velocity of the ball just before it hits the inclined plane.
The ball falls freely for 2 seconds under the influence of gravity. The equation for velocity under constant acceleration is:
\[
v = u + gt
\]
Where:
- \( u = 0 \) (initial velocity)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 2 \, \text{s} \)
Substituting the values:
\[
v = 0 + (10 \, \text{m/s}^2)(2 \, \text{s}) = 20 \, \text{m/s}
\]
### Step 2: Resolve the velocity into components along and perpendicular to the inclined plane.
The inclined plane makes an angle of \( 30^\circ \) with the horizontal. We need to resolve the velocity into two components:
1. Perpendicular to the plane
2. Along the plane
Using trigonometry:
- The component of velocity perpendicular to the plane (\( v_{\perp} \)) is given by:
\[
v_{\perp} = v \cos(30^\circ) = 20 \cos(30^\circ) = 20 \left(\frac{\sqrt{3}}{2}\right) = 10\sqrt{3} \, \text{m/s}
\]
- The component of velocity along the plane (\( v_{\parallel} \)) is given by:
\[
v_{\parallel} = v \sin(30^\circ) = 20 \sin(30^\circ) = 20 \left(\frac{1}{2}\right) = 10 \, \text{m/s}
\]
### Step 3: Calculate the velocity after the first impact using the coefficient of restitution.
The coefficient of restitution \( e \) is given as \( \frac{5}{8} \). The relationship between the velocities before and after the impact is given by:
\[
e = \frac{\text{velocity of separation}}{\text{velocity of approach}}
\]
The velocity of approach is the perpendicular component:
\[
\text{velocity of approach} = v_{\perp} = 10\sqrt{3} \, \text{m/s}
\]
The velocity of separation can be calculated as:
\[
\text{velocity of separation} = e \cdot \text{velocity of approach} = \frac{5}{8} \cdot 10\sqrt{3} = \frac{50\sqrt{3}}{8} = \frac{25\sqrt{3}}{4} \, \text{m/s}
\]
### Step 4: Determine the new vertical component of velocity after impact.
The new vertical component of velocity after impact is:
\[
v'_{\perp} = v_{\perp} - \text{velocity of separation} = 10\sqrt{3} - \frac{25\sqrt{3}}{4}
\]
Finding a common denominator:
\[
v'_{\perp} = \frac{40\sqrt{3}}{4} - \frac{25\sqrt{3}}{4} = \frac{15\sqrt{3}}{4} \, \text{m/s}
\]
### Step 5: Calculate the time of flight after the first impact.
The acceleration component perpendicular to the inclined plane is:
\[
a_{\perp} = g \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s}^2
\]
Using the formula for time of flight \( t \):
\[
t = \frac{v'_{\perp}}{a_{\perp}} = \frac{\frac{15\sqrt{3}}{4}}{5\sqrt{3}} = \frac{15}{20} = \frac{3}{4} \, \text{s}
\]
### Step 6: Calculate the distance traveled along the inclined plane during this time.
The distance along the inclined plane can be calculated using the formula:
\[
s = v_{\parallel} t + \frac{1}{2} a_{\parallel} t^2
\]
Where \( a_{\parallel} = g \sin(30^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s}^2 \).
Substituting the values:
\[
s = (10 \, \text{m/s})(\frac{3}{4} \, \text{s}) + \frac{1}{2}(5 \, \text{m/s}^2)(\frac{3}{4} \, \text{s})^2
\]
Calculating each term:
1. First term: \( 10 \cdot \frac{3}{4} = 7.5 \, \text{m} \)
2. Second term: \( \frac{1}{2} \cdot 5 \cdot \frac{9}{16} = \frac{45}{16} = 2.8125 \, \text{m} \)
Adding both terms:
\[
s = 7.5 + 2.8125 = 10.3125 \, \text{m}
\]
### Final Result
The distance along the plane between the first and second impact of the ball is approximately \( 10.3125 \, \text{m} \).
