A projectile is fired at an angle of `30^(@)` with the horizontal such that the vertical component of its initial velocity is 80m/s. Find approximatly the velocity of the projectile at time `T//4` where T is time of flight.

Options

A. `180m//s`
B. `155m//s`
C. `145m//s`
D. `140m//s`

Explanation

To solve the problem step by step, we will follow the concepts of projectile motion.

### Step 1: Determine the initial velocity (u)
Given that the vertical component of the initial velocity \( u_y = 80 \, \text{m/s} \) and the angle of projection \( \theta = 30^\circ \), we can use the relationship between the vertical component and the initial velocity:
\[ u_y = u \sin \theta \]
Substituting the known values:
\[ 80 = u \sin(30^\circ) \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ 80 = u \cdot \frac{1}{2} \]
Thus, solving for \( u \):
\[ u = 80 \cdot 2 = 160 \, \text{m/s} \]

### Step 2: Calculate the horizontal component of the initial velocity (\( u_x \))
Using the initial velocity \( u \) and the angle \( \theta \):
\[ u_x = u \cos \theta \]
Substituting the known values:
\[ u_x = 160 \cos(30^\circ) \]
Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ u_x = 160 \cdot \frac{\sqrt{3}}{2} = 80\sqrt{3} \, \text{m/s} \]

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

### Step 4: Determine the time \( t = \frac{T}{4} \)
Calculating \( t \):
\[ t = \frac{16}{4} = 4 \, \text{s} \]

### Step 5: Calculate the vertical component of velocity at time \( t \) (\( v_y \))
The vertical velocity at time \( t \) is given by:
\[ v_y = u_y - g t \]
Substituting the known values:
\[ v_y = 80 - 10 \cdot 4 = 80 - 40 = 40 \, \text{m/s} \]

### Step 6: The horizontal component of velocity (\( v_x \))
The horizontal component of velocity remains constant:
\[ v_x = u_x = 80\sqrt{3} \, \text{m/s} \]

### Step 7: Calculate the resultant velocity (\( v \))
The magnitude of the resultant velocity is given by:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(80\sqrt{3})^2 + (40)^2} \]
Calculating:
\[ v = \sqrt{(6400 \cdot 3) + 1600} = \sqrt{19200 + 1600} = \sqrt{20800} \]
\[ v \approx 144.22 \, \text{m/s} \]

### Final Answer
The approximate velocity of the projectile at time \( \frac{T}{4} \) is \( \approx 145 \, \text{m/s} \).

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