A projectile is fired at an angle of `30^(@)` to the horizontal such that the vertical component of its initial velocity is `80m//s`. Its time of fight is `T`. Its velocity at `t=T//4` has a magnitude of nearly.

Options

A. `200 ms^(-1)`
B. `300 ms^(-1)`
C. `100 ms^(-1)`
D. None of these

Explanation

To solve the problem step by step, we will calculate the velocity of the projectile at \( t = \frac{T}{4} \).

### Step 1: Identify the given information
- The angle of projection \( \theta = 30^\circ \)
- The vertical component of the initial velocity \( u_y = 80 \, \text{m/s} \)
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the horizontal component of the initial velocity
Using the relationship between the vertical and horizontal components of the initial velocity:
\[ u_y = u \sin \theta \]
\[ u_x = u \cos \theta \]
From the problem, we know:
\[ u_y = 80 \, \text{m/s} \quad \text{and} \quad \sin 30^\circ = \frac{1}{2} \]
Thus,
\[ u = \frac{u_y}{\sin 30^\circ} = \frac{80}{\frac{1}{2}} = 160 \, \text{m/s} \]
Now, calculate \( u_x \):
\[ u_x = u \cos 30^\circ = 160 \cdot \frac{\sqrt{3}}{2} = 80\sqrt{3} \, \text{m/s} \]

### Step 3: Calculate the time of flight \( T \)
The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u_y}{g} \]
Substituting the known values:
\[ T = \frac{2 \cdot 80}{10} = 16 \, \text{s} \]

### Step 4: Calculate \( t = \frac{T}{4} \)
Now, we find \( t \):
\[ t = \frac{T}{4} = \frac{16}{4} = 4 \, \text{s} \]

### Step 5: Calculate the horizontal and vertical components of velocity at \( t = 4 \, \text{s} \)
The horizontal component of velocity remains constant:
\[ v_x = u_x = 80\sqrt{3} \, \text{m/s} \]

For the vertical component of velocity, we use the formula:
\[ v_y = u_y - g \cdot t \]
Substituting the values:
\[ v_y = 80 - 10 \cdot 4 = 80 - 40 = 40 \, \text{m/s} \]

### Step 6: Calculate the magnitude of the resultant velocity
The magnitude of the resultant velocity \( v \) is given by:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(80\sqrt{3})^2 + (40)^2} = \sqrt{19200 + 1600} = \sqrt{20800} \]
Calculating further:
\[ v = \sqrt{20800} = \sqrt{16 \cdot 1300} = 4\sqrt{1300} \approx 4 \cdot 36.06 \approx 144.24 \, \text{m/s} \]

### Final Answer
The magnitude of the velocity at \( t = \frac{T}{4} \) is approximately \( 144.24 \, \text{m/s} \).

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