An arrow is projected into air. It's time of flight is 5seconds and range 200m. , the maximum height attained by the arrow is:-

Options

A. 25m
B. 40m
C. 31.25m
D. 12.5m

Explanation

To find the maximum height attained by the arrow, we can use the information provided about the time of flight and the range. Here’s the step-by-step solution:

### Step 1: Understand the given data
- Time of flight (T) = 5 seconds
- Range (R) = 200 meters
- Acceleration due to gravity (g) = 10 m/s² (approximate value)

### Step 2: Calculate the vertical component of the initial velocity (u_y)
The total time of flight for a projectile is given by the formula:
\[ T = \frac{2u_y}{g} \]
Where:
- \( u_y \) is the vertical component of the initial velocity.

Rearranging the formula to find \( u_y \):
\[ u_y = \frac{gT}{2} \]

Substituting the known values:
\[ u_y = \frac{10 \times 5}{2} = 25 \, \text{m/s} \]

### Step 3: Calculate the maximum height (h_max)
The maximum height attained by a projectile is given by the formula:
\[ h_{max} = \frac{u_y^2}{2g} \]

Substituting the value of \( u_y \) we found:
\[ h_{max} = \frac{(25)^2}{2 \times 10} \]
\[ h_{max} = \frac{625}{20} = 31.25 \, \text{m} \]

### Step 4: Conclusion
The maximum height attained by the arrow is **31.25 meters**.