Consider a rubber ball freely falling from a height ` h = 4.9 m` onto a horizontally elastic plate. Assume that the duration of collision is negligible and the collisions 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

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Answer

To solve the problem of a rubber ball falling from a height of \( h = 4.9 \, \text{m} \) onto a horizontally elastic plate, we need to determine the velocity and height of the ball as functions of time.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The ball is dropped from rest, so its initial velocity \( u = 0 \).
- The height from which it is dropped is \( h = 4.9 \, \text{m} \).

2. **Using the Kinematic Equation for Velocity**:
- The equation for the velocity of an object under free fall is given by:
\[ v = u + at \]
- Here, \( a = -g \) (acceleration due to gravity, negative because it is directed downwards), and \( g \approx 9.8 \, \text{m/s}^2 \).
- Substituting \( u = 0 \):
\[ v = 0 - gt = -gt \]
- Therefore, the velocity as a function of time is:
\[ v(t) = -9.8t \quad \text{(for the downward motion)} \]

3. **Calculating the Time of Fall**:
- To find the time \( t \) it takes to fall \( 4.9 \, \text{m} \), we use the second kinematic equation:
\[ s = ut + \frac{1}{2}at^2 \]
- Here, \( s = -4.9 \, \text{m} \) (since it falls down), \( u = 0 \), and \( a = -g \):
\[ -4.9 = 0 + \frac{1}{2}(-9.8)t^2 \]
- Simplifying gives:
\[ -4.9 = -4.9t^2 \quad \Rightarrow \quad t^2 = 1 \quad \Rightarrow \quad t = 1 \, \text{s} \]

4. **Behavior During the Collision**:
- Since the collision with the elastic plate is perfectly elastic, the ball will rebound with the same speed but in the opposite direction.
- Therefore, after \( t = 1 \, \text{s} \), the velocity will be:
\[ v = 9.8 \, \text{m/s} \quad \text{(upward)} \]

5. **Height as a Function of Time**:
- For the upward motion after the collision, we again use the kinematic equation:
\[ s = ut + \frac{1}{2}at^2 \]
- For the upward motion, the initial velocity \( u = 9.8 \, \text{m/s} \) and \( a = -g \):
\[ h(t) = 4.9 + 9.8t - \frac{1}{2}(9.8)t^2 \]
- This will continue until the ball reaches the maximum height and then falls back down.

6. **Generalizing the Motion**:
- The ball will continue to bounce indefinitely, following the same pattern of falling and rebounding, maintaining the same height of \( 4.9 \, \text{m} \) for each bounce.

### Final Functions:
- **Velocity as a function of time**:
\[ v(t) = \begin{cases} -9.8t & \text{for } 0 \leq t < 1 \\ 9.8(t - 2n) & \text{for } t \text{ in the } n^{th} \text{ bounce} \end{cases} \]
- **Height as a function of time**:
\[ h(t) = \begin{cases} 4.9 - \frac{1}{2}gt^2 & \text{for } 0 \leq t < 1 \\ 4.9 + 9.8(t - 2n) - \frac{1}{2}g(t - 2n)^2 & \text{for } t \text{ in the } n^{th} \text{ bounce} \end{cases} \]