In the javelin throw at a track-and-field event, the javelin is launched at a speed of 29 m/s at an angle of `36^(@)` above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from `36^(@)` at launch to `18^(@)`?

Options

A. 0.96 s
B. 1.12 s
C. 1.04 s
D. 1.16 s

Answer

To solve the problem of how much time is required for the angle of the javelin's velocity to decrease from \(36^\circ\) to \(18^\circ\), we can follow these steps:

### Step 1: Determine the initial velocity components
The javelin is launched at a speed of \(29 \, \text{m/s}\) at an angle of \(36^\circ\). We can find the horizontal and vertical components of the initial velocity using trigonometric functions.

- **Horizontal component (\(V_{x}\))**:
\[ V_{x} = V \cdot \cos(\theta) = 29 \cdot \cos(36^\circ) \]
Calculating this gives:
\[ V_{x} \approx 29 \cdot 0.809 = 23.46 \, \text{m/s} \]

- **Vertical component (\(V_{y}\))**:
\[ V_{y} = V \cdot \sin(\theta) = 29 \cdot \sin(36^\circ) \]
Calculating this gives:
\[ V_{y} \approx 29 \cdot 0.588 = 17.03 \, \text{m/s} \]

### Step 2: Determine the vertical velocity when the angle is \(18^\circ\)
When the angle of the javelin's velocity is \(18^\circ\), we can use the tangent function to find the vertical velocity (\(V_{y}\)) at that angle.

Using the relationship:
\[ \tan(18^\circ) = \frac{V_{y}}{V_{x}} \]
We can rearrange this to find \(V_{y}\):
\[ V_{y} = V_{x} \cdot \tan(18^\circ) = 23.46 \cdot \tan(18^\circ) \]
Calculating this gives:
\[ V_{y} \approx 23.46 \cdot 0.3249 \approx 7.62 \, \text{m/s} \]

### Step 3: Use the kinematic equation to find the time
We can use the kinematic equation for vertical motion to find the time required for the vertical velocity to decrease from \(17.03 \, \text{m/s}\) to \(7.62 \, \text{m/s}\):
\[ V_{y} = V_{y0} - g \cdot t \]
Where:
- \(V_{y} = 7.62 \, \text{m/s}\) (final vertical velocity)
- \(V_{y0} = 17.03 \, \text{m/s}\) (initial vertical velocity)
- \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity)

Substituting the values:
\[ 7.62 = 17.03 - 9.8 \cdot t \]
Rearranging gives:
\[ 9.8 \cdot t = 17.03 - 7.62 \]
\[ 9.8 \cdot t = 9.41 \]
\[ t = \frac{9.41}{9.8} \approx 0.96 \, \text{s} \]

### Final Answer
The time required for the angle to be reduced from \(36^\circ\) to \(18^\circ\) is approximately \(0.96\) seconds.

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