A ball is dropped from a height of `320 m` above the ground. After every collision, the speed of ball decreases by `50%`. Taking dropping point as origin, downward direction positive and collision time negligible, sketch the v-t, s-t and a-t graphs. Also calculate the total distance traveled by the ball and the total time of journey.

Options

A. `omega_(c)-omega_(m)`
B. `omega_(m)`
C. `omega_(c)`
D. `omega_(m)+omega_(c)`

Answer

To solve the problem step by step, we will analyze the motion of the ball dropped from a height of 320 meters, considering the effects of collisions that reduce its speed by 50% after each bounce.

### Step 1: Calculate the initial velocity just before the first collision
The ball is dropped from a height \( h = 320 \, m \). Using the equation of motion:
\[ v^2 = u^2 + 2gh \]
where:
- \( u = 0 \) (initial velocity),
- \( g = 10 \, m/s^2 \) (acceleration due to gravity),
- \( h = 320 \, m \).

Substituting the values:
\[ v^2 = 0 + 2 \times 10 \times 320 \]
\[ v^2 = 6400 \implies v = \sqrt{6400} = 80 \, m/s \]

### Step 2: Calculate the height after the first collision
After the first collision, the speed of the ball decreases by 50%, so:
\[ v_1 = \frac{v}{2} = \frac{80}{2} = 40 \, m/s \]
Now, we calculate the height \( h_1 \) it reaches after the collision using:
\[ v^2 = u^2 + 2gh_1 \]
Here, \( v = 0 \) (final velocity at the peak), \( u = 40 \, m/s \), and \( g = 10 \, m/s^2 \):
\[ 0 = (40)^2 - 2 \times 10 \times h_1 \]
\[ 1600 = 20h_1 \implies h_1 = \frac{1600}{20} = 80 \, m \]

### Step 3: Calculate the height after the second collision
After the second collision, the speed reduces again by 50%:
\[ v_2 = \frac{v_1}{2} = \frac{40}{2} = 20 \, m/s \]
Calculating the height \( h_2 \):
\[ 0 = (20)^2 - 2 \times 10 \times h_2 \]
\[ 400 = 20h_2 \implies h_2 = \frac{400}{20} = 20 \, m \]

### Step 4: Calculate the height after subsequent collisions
Continuing this pattern, we find that:
- \( h_3 = \frac{h_2}{4} = \frac{20}{4} = 5 \, m \)
- \( h_4 = \frac{h_3}{4} = \frac{5}{4} = 1.25 \, m \)
- And so on...

### Step 5: Total distance traveled
The total distance \( D \) can be calculated as:
\[ D = 320 + 2h_1 + 2h_2 + 2h_3 + 2h_4 + \ldots \]
This series can be expressed as:
\[ D = 320 + 2 \left( h_1 + h_2 + h_3 + h_4 + \ldots \right) \]
The heights form a geometric series:
\[ h_1 + h_2 + h_3 + h_4 + \ldots = 80 + 20 + 5 + 1.25 + \ldots \]
This series has a first term \( a = 80 \) and a common ratio \( r = \frac{1}{4} \):
\[ S = \frac{a}{1 - r} = \frac{80}{1 - \frac{1}{4}} = \frac{80}{\frac{3}{4}} = \frac{320}{3} \]
Thus:
\[ D = 320 + 2 \times \frac{320}{3} = 320 + \frac{640}{3} = \frac{960 + 640}{3} = \frac{1600}{3} \, m \]

### Step 6: Total time of journey
The time taken for each segment can be calculated as follows:
- For the first fall (320 m):
\[ t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 320}{10}} = \sqrt{64} = 8 \, s \]
- For the first rise (80 m):
\[ t_2 = \frac{v_1}{g} = \frac{40}{10} = 4 \, s \]
- For the second fall (80 m):
\[ t_3 = 4 \, s \]
- For the second rise (20 m):
\[ t_4 = \frac{20}{10} = 2 \, s \]
Continuing this pattern, the total time can be calculated as:
\[ T = 8 + 4 + 4 + 2 + 2 + 1 + 0.5 + \ldots \]
This series also forms a geometric series:
\[ T = 8 + 8 + 4 + 2 + 1 + 0.5 + \ldots \]
The sum of the series is:
\[ T = 8 + 8 \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \right) = 8 + 8 \left( \frac{\frac{1}{2}}{1 - \frac{1}{2}} \right) = 8 + 8 = 24 \, s \]

### Final Results
- Total distance traveled by the ball: \( \frac{1600}{3} \, m \)
- Total time of journey: \( 24 \, s \)