The relation between the time of flight of projectile `T_(f)` and the time to reach the maximum height `t_(m)` is

Options

A. `T_(f) = 2t_(m)`
B. `T_(f) = t_(m)`
C. `T_(f) = (t_(m))/(2)`
D. `T_(f) = sqrt(2)(t_(m))`

Explanation

To establish the relation between the time of flight of a projectile \( T_f \) and the time to reach the maximum height \( t_m \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion of the Projectile**:
- When a projectile is launched at an angle \( \theta \) with an initial velocity \( u \), it follows a parabolic trajectory.
- The motion can be divided into two parts: the ascent (going up to the maximum height) and the descent (falling back down).

2. **Time to Reach Maximum Height**:
- The time taken to reach the maximum height \( t_m \) can be derived from the equations of motion. At maximum height, the vertical component of the velocity becomes zero.
- The vertical component of the initial velocity is given by \( u_y = u \sin \theta \).
- Using the equation \( v = u + at \) (where \( v = 0 \) at maximum height and \( a = -g \)), we have:
\[ 0 = u \sin \theta - g t_m \]
- Rearranging gives:
\[ t_m = \frac{u \sin \theta}{g} \]

3. **Total Time of Flight**:
- The total time of flight \( T_f \) is the time taken to go up to the maximum height and then return back down to the original launch height.
- Since the time taken to ascend to maximum height is equal to the time taken to descend back to the original height, we have:
\[ T_f = t_m + t_m = 2t_m \]

4. **Final Relation**:
- Substituting the expression for \( t_m \) into the equation for \( T_f \):
\[ T_f = 2 \left( \frac{u \sin \theta}{g} \right) \]
- Thus, the relation between the time of flight \( T_f \) and the time to reach maximum height \( t_m \) is:
\[ T_f = 2t_m \]

### Conclusion:
The relation between the time of flight \( T_f \) and the time to reach maximum height \( t_m \) is given by:
\[ T_f = 2t_m \]