A projectile is thrown with initial velocity `u_(0)` and angle `30^(@)` with the horizontal. If it remains in the air for 1s. What was its initial velocity ?

Options

A. `19.6 ms^(-1)`
B. `9.8 ms^(-1)`
C. `4.9 ms^(-1)`
D. `1ms^(-1)`

Explanation

To solve the problem of finding the initial velocity \( u_0 \) of a projectile thrown at an angle of \( 30^\circ \) with the horizontal, and which remains in the air for 1 second, we can follow these steps:

### Step 1: Understand the Time of Flight Formula
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
where:
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 2: Substitute the Given Values
In this problem, we know:
- \( T = 1 \, \text{s} \)
- \( \theta = 30^\circ \)

Substituting these values into the time of flight formula:
\[ 1 = \frac{2u_0 \sin(30^\circ)}{g} \]

### Step 3: Calculate \( \sin(30^\circ) \)
We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]

### Step 4: Substitute \( \sin(30^\circ) \) into the Equation
Now substituting \( \sin(30^\circ) \) into the equation:
\[ 1 = \frac{2u_0 \cdot \frac{1}{2}}{g} \]

### Step 5: Simplify the Equation
This simplifies to:
\[ 1 = \frac{u_0}{g} \]

### Step 6: Solve for \( u_0 \)
To find \( u_0 \), we rearrange the equation:
\[ u_0 = g \]

### Step 7: Substitute the Value of \( g \)
Using \( g \approx 9.8 \, \text{m/s}^2 \):
\[ u_0 = 9.8 \, \text{m/s}^2 \]

### Conclusion
Thus, the initial velocity \( u_0 \) of the projectile is:
\[ \boxed{9.8 \, \text{m/s}} \]