A body projected at an angle with the horizontal has a range 300 m. the time of flight is 6s, then the horizontal component of velocity is

Options

A. 30 m/s
B. 50 m/s
C. 40 m/s
D. 45 m/s

Explanation

To find the horizontal component of velocity for a body projected at an angle with a given range and time of flight, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Range (R) = 300 m
- Time of Flight (T) = 6 s

2. **Recall the Formula for Range in Projectile Motion:**
The formula for the range (R) of a projectile is given by:
\[ R = u \cos \theta \cdot T \]
where \( u \) is the initial velocity and \( \theta \) is the angle of projection.

3. **Rearranging the Formula:**
We can rearrange the formula to find the horizontal component of velocity (\( u \cos \theta \)):
\[ u \cos \theta = \frac{R}{T} \]

4. **Substituting the Known Values:**
Now, substitute the known values of range and time of flight into the equation:
\[ u \cos \theta = \frac{300 \, \text{m}}{6 \, \text{s}} = 50 \, \text{m/s} \]

5. **Conclusion:**
Therefore, the horizontal component of velocity is:
\[ u \cos \theta = 50 \, \text{m/s} \]

### Final Answer:
The horizontal component of velocity is **50 m/s**.

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