A javelin thrown into air at an angle with the horizontal has a range of 200 m. If the time of flight is 5 second, then the horizontal component of velocity of the projectile at the highest point of the trajectory is

Options

A. `40 m//s`
B. `20 m//s`
C. `9.8 m//s`
D. `5m//s`

Answer

To solve the problem of finding the horizontal component of velocity of a javelin thrown into the air, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
We are given the range of the projectile (javelin) as 200 meters and the total time of flight as 5 seconds. We need to find the horizontal component of velocity at the highest point of the trajectory.

2. **Identify the Components of Motion**:
In projectile motion, the motion can be broken down into horizontal and vertical components. The horizontal component of velocity remains constant throughout the flight because there is no horizontal acceleration (assuming air resistance is negligible).

3. **Use the Formula for Horizontal Velocity**:
The horizontal component of velocity (u_x) can be calculated using the formula:
\[ u_x = \frac{\text{Range}}{\text{Time of Flight}} \]

4. **Substitute the Given Values**:
Substitute the given values into the formula:
\[ u_x = \frac{200 \, \text{m}}{5 \, \text{s}} \]

5. **Calculate the Horizontal Component of Velocity**:
Performing the calculation:
\[ u_x = 40 \, \text{m/s} \]

6. **Conclusion**:
The horizontal component of velocity of the projectile at the highest point of the trajectory is **40 m/s**.

### Final Answer:
The horizontal component of velocity of the javelin at the highest point of the trajectory is **40 m/s**.

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