A ball is dropped from a platform 19.6m high. Its position function is

Options

A. `x= -4.9 t^(2) + 19.6 (0 le t le 1)`
B. `x = -4.9 t^(2) + 19.6 (0 le t le 2)`
C. `x= - 9.8 t^(2) + 19.6 (0 le t le 2)`
D. `x= -4.9 + 19.6 (0 le t le 2)`

Answer

To find the position function of a ball dropped from a height of 19.6 meters, we can use the equations of motion under constant acceleration due to gravity. Here’s a step-by-step solution:

### Step 1: Understand the problem
The ball is dropped from a height of 19.6 meters. We need to establish the position function of the ball as it falls under the influence of gravity.

### Step 2: Identify the known values
- Initial height (h) = 19.6 m
- Initial velocity (u) = 0 m/s (since the ball is dropped)
- Acceleration (a) = g = 9.8 m/s² (acceleration due to gravity)

### Step 3: Use the position function formula
The position function \( s(t) \) for an object under constant acceleration can be expressed as:
\[ s(t) = ut + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ s(t) = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
This simplifies to:
\[ s(t) = 4.9 t^2 \]

### Step 4: Determine the position relative to the height
Since the ball is falling from a height of 19.6 meters, we need to express the position function in terms of the height above the ground:
\[ \text{Height above ground} = \text{Initial height} - \text{Distance fallen} \]
Thus, we can write:
\[ h(t) = 19.6 - s(t) \]
Substituting \( s(t) \):
\[ h(t) = 19.6 - 4.9 t^2 \]

### Step 5: Define the time interval
The ball will hit the ground when \( h(t) = 0 \):
\[ 19.6 - 4.9 t^2 = 0 \]
Solving for \( t \):
\[ 4.9 t^2 = 19.6 \]
\[ t^2 = \frac{19.6}{4.9} = 4 \]
\[ t = 2 \text{ seconds} \]
Thus, the ball will be in motion from \( t = 0 \) to \( t = 2 \) seconds.

### Final Position Function
The position function of the ball as it falls from the platform is:
\[ h(t) = 19.6 - 4.9 t^2 \quad \text{for } 0 \leq t \leq 2 \]