Consider a rubber ball freely falling from a height h = 4.9 m on a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as a function of time will be:

Options

A.
B.
C.
D.

Explanation

To solve the problem of a rubber ball falling from a height of 4.9 m onto a horizontal elastic plate, we need to determine the velocity and height of the ball as functions of time. We will consider the motion of the ball before and after the collision with the plate.

### Step-by-Step Solution:

1. **Determine the time taken to fall to the plate:**
The ball is dropped from a height \( h = 4.9 \, \text{m} \). Using the equation of motion:
\[ h = \frac{1}{2} g t^2 \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)). Rearranging gives:
\[ t^2 = \frac{2h}{g} = \frac{2 \times 4.9}{9.8} = 1 \implies t = 1 \, \text{s} \]

**Hint:** Use the equation of motion for free fall to find the time taken to reach the ground.

2. **Calculate the velocity just before impact:**
The velocity of the ball just before it hits the plate can be calculated using:
\[ v = gt \]
Substituting \( t = 1 \, \text{s} \):
\[ v = 9.8 \times 1 = 9.8 \, \text{m/s} \]

**Hint:** Use the formula for velocity under constant acceleration to find the speed just before impact.

3. **Analyze the collision:**
Since the collision is perfectly elastic, the ball will rebound with the same speed but in the opposite direction. Thus, the velocity after the collision will be:
\[ v_{\text{after}} = -9.8 \, \text{m/s} \]

**Hint:** Remember that in a perfectly elastic collision, the speed remains the same, but the direction reverses.

4. **Determine the height as a function of time after the collision:**
After the collision, the ball will ascend back to the height of 4.9 m. The time taken to reach the maximum height can be found using:
\[ v = u + at \]
where \( u = -9.8 \, \text{m/s} \) (upward), \( v = 0 \) (at the maximum height), and \( a = -g \):
\[ 0 = -9.8 - 9.8t \implies t = 1 \, \text{s} \]
Thus, it takes another 1 second to reach the maximum height.

5. **Height as a function of time:**
The height \( h(t) \) can be modeled as:
- For \( 0 \leq t < 1 \):
\[ h(t) = 4.9 - \frac{1}{2} g t^2 \]
- For \( 1 \leq t < 2 \):
\[ h(t) = 4.9 + \frac{1}{2} g (t - 1)^2 \]

**Hint:** Use the equations of motion to describe the height during both the falling and rising phases.

6. **Velocity as a function of time:**
The velocity \( v(t) \) can be described as:
- For \( 0 \leq t < 1 \):
\[ v(t) = -gt \]
- For \( 1 \leq t < 2 \):
\[ v(t) = -9.8 + g(t - 1) \]

**Hint:** Use the equations of motion to describe the velocity during both the falling and rising phases.

### Summary:
- **Velocity as a function of time:**
\[ v(t) = \begin{cases} -9.8t & \text{for } 0 \leq t < 1 \\ 9.8(t - 1) & \text{for } 1 \leq t < 2 \end{cases} \]

- **Height as a function of time:**
\[ h(t) = \begin{cases} 4.9 - \frac{1}{2} g t^2 & \text{for } 0 \leq t < 1 \\ 4.9 + \frac{1}{2} g (t - 1)^2 & \text{for } 1 \leq t < 2 \end{cases} \]