In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.050 s, during which time it experiences an acceleration of `340m//s^(2)`. The ball is launched at an angle of `51^(@)` above the ground. Determine the horizontal and vertical components of the launch velocity.

Options

A. `{:(v_(x),v_(y)),(13m//s,11m//s):}`
B. `{:(v_(x),v_(y)),(21m//s,11m//s):}`
C. `{:(v_(x),v_(y)),(11m//s,13m//s):}`
D. `{:(v_(x),v_(y)),(17m//s,21m//s):}`

Answer

To solve the problem, we will break it down into steps:

### Step 1: Calculate the final velocity of the ball
We know that the ball experiences a constant acceleration while in contact with the kicker's foot. We can use the formula for final velocity under constant acceleration:

\[ v = u + at \]

Where:
- \( v \) = final velocity
- \( u \) = initial velocity (which is 0 since the ball starts from rest)
- \( a \) = acceleration (given as \( 340 \, \text{m/s}^2 \))
- \( t \) = time of contact (given as \( 0.050 \, \text{s} \))

Substituting the values:

\[ v = 0 + (340 \, \text{m/s}^2)(0.050 \, \text{s}) = 17 \, \text{m/s} \]

### Step 2: Resolve the final velocity into horizontal and vertical components
The final velocity of the ball can be resolved into horizontal and vertical components using the angle of launch. The angle given is \( 51^\circ \).

The horizontal component \( v_x \) is given by:

\[ v_x = v \cdot \cos(\theta) \]

And the vertical component \( v_y \) is given by:

\[ v_y = v \cdot \sin(\theta) \]

Where:
- \( \theta = 51^\circ \)
- \( v = 17 \, \text{m/s} \)

Calculating \( v_x \):

\[ v_x = 17 \cdot \cos(51^\circ) \approx 17 \cdot 0.6293 \approx 10.69 \, \text{m/s} \]

Calculating \( v_y \):

\[ v_y = 17 \cdot \sin(51^\circ) \approx 17 \cdot 0.7771 \approx 13.21 \, \text{m/s} \]

### Step 3: State the final components of the launch velocity
Thus, the horizontal and vertical components of the launch velocity are:

- Horizontal component \( v_x \approx 10.69 \, \text{m/s} \)
- Vertical component \( v_y \approx 13.21 \, \text{m/s} \)

### Final Answer
- Horizontal component \( v_x \approx 10.69 \, \text{m/s} \)
- Vertical component \( v_y \approx 13.21 \, \text{m/s} \)

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