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 find the horizontal component of velocity of the javelin at the highest point of its trajectory, we can follow these steps:

### Step 1: Understand the given data
We are given:
- Range (R) = 200 m
- Time of flight (T) = 5 s

### Step 2: Recall the formula for range
The formula for the range of a projectile is given by:
\[ R = v \cdot \cos(\theta) \cdot T \]
where:
- \( R \) is the range,
- \( v \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( T \) is the time of flight.

### Step 3: Rearranging the formula
We need to find the horizontal component of velocity at the highest point, which is equal to \( v \cdot \cos(\theta) \). Rearranging the formula for range gives us:
\[ v \cdot \cos(\theta) = \frac{R}{T} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ v \cdot \cos(\theta) = \frac{200 \, \text{m}}{5 \, \text{s}} \]

### Step 5: Calculate the horizontal component of velocity
Now, performing the calculation:
\[ v \cdot \cos(\theta) = \frac{200}{5} = 40 \, \text{m/s} \]

### Conclusion
Thus, the horizontal component of velocity of the projectile at the highest point of the trajectory is:
\[ \text{Horizontal component of velocity} = 40 \, \text{m/s} \]