When a particle is thrown horizontally, the resultant velocity of the projectile at any time t is given by:-

Options

A. gt
B. `(1)/(2)"gt"^(2)`
C. `sqrt(u^(2)+g^(2)t^(2))`
D. `sqrt(u^(2)-g^(2)t^(2))`

Explanation

To find the resultant velocity of a projectile when a particle is thrown horizontally, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components of Velocity:**
- When a particle is thrown horizontally, it has two components of velocity:
- Horizontal component (\(v_x\))
- Vertical component (\(v_y\))

2. **Determine the Horizontal Component:**
- The horizontal component of velocity remains constant throughout the motion since there are no horizontal forces acting on the particle (assuming air resistance is negligible).
- Let the initial horizontal velocity be \(u\). Therefore,
\[ v_x = u \]

3. **Determine the Vertical Component:**
- The vertical component of velocity changes due to the acceleration caused by gravity.
- The initial vertical velocity is \(0\) (since it is thrown horizontally).
- The vertical component of velocity at time \(t\) can be calculated using the equation:
\[ v_y = u_y + gt \]
where \(u_y = 0\) (initial vertical velocity) and \(g\) is the acceleration due to gravity.
- Thus,
\[ v_y = 0 + gt = gt \]

4. **Calculate the Resultant Velocity:**
- The resultant velocity (\(v\)) of the projectile can be found using the Pythagorean theorem, since the horizontal and vertical components are perpendicular to each other:
\[ v = \sqrt{v_x^2 + v_y^2} \]
- Substituting the values of \(v_x\) and \(v_y\):
\[ v = \sqrt{u^2 + (gt)^2} \]

5. **Final Result:**
- Therefore, the resultant velocity of the projectile at any time \(t\) is given by:
\[ v = \sqrt{u^2 + (gt)^2} \]