If g is acceleration due to gravity , then the vertical distance travelled by a projectile in time t is equal to

Options

A. `(1)/ (2)g t^(2)`
B. gt
C. `(3)/(4)gh^(2)`
D. none of these

Explanation

To find the vertical distance traveled by a projectile in time \( t \), we can use the equations of motion. Here's the step-by-step solution:

### Step 1: Understand the Components of Motion
When a projectile is launched, it has both horizontal and vertical components of motion. The vertical motion is influenced by gravity.

### Step 2: Identify Initial Conditions
1. **Initial Velocity in the Vertical Direction (\( u_y \))**: If the projectile is launched at an angle, the vertical component of the initial velocity is \( u \sin \theta \). However, if we are considering the vertical distance traveled from the highest point downwards, we can assume the initial vertical velocity is \( 0 \) when calculating the distance fallen under gravity.
2. **Acceleration (\( a \))**: The acceleration due to gravity is \( g \), acting downwards.

### Step 3: Use the Equation of Motion
We can use the second equation of motion to find the vertical distance (\( S_y \)) traveled in time \( t \):
\[ S_y = u_y t + \frac{1}{2} a t^2 \]
Substituting the values:
- \( u_y = 0 \) (initial vertical velocity)
- \( a = g \) (acceleration due to gravity)

### Step 4: Substitute Values into the Equation
Substituting the values into the equation:
\[ S_y = 0 \cdot t + \frac{1}{2} g t^2 \]
This simplifies to:
\[ S_y = \frac{1}{2} g t^2 \]

### Step 5: Conclusion
Thus, the vertical distance traveled by the projectile in time \( t \) is:
\[ S_y = \frac{1}{2} g t^2 \]