A body is projected at an angle of `30^(@)` with a velocity of `320 ft//"sec"`, then
the direction of motion of velocity after the lapse of two seconds ..........

Answer

To solve the problem of finding the direction of motion of a projectile after 2 seconds, we will follow these steps:

### Step 1: Break down the initial velocity into components
The initial velocity \( v_0 \) is given as \( 320 \, \text{ft/s} \) at an angle of \( 30^\circ \). We can find the horizontal and vertical components of the velocity using trigonometric functions.

- **Horizontal Component** (\( v_{x0} \)):
\[ v_{x0} = v_0 \cos(30^\circ) = 320 \cos(30^\circ) = 320 \times \frac{\sqrt{3}}{2} = 160\sqrt{3} \, \text{ft/s} \]

- **Vertical Component** (\( v_{y0} \)):
\[ v_{y0} = v_0 \sin(30^\circ) = 320 \sin(30^\circ) = 320 \times \frac{1}{2} = 160 \, \text{ft/s} \]

### Step 2: Calculate the horizontal velocity after 2 seconds
The horizontal component of the velocity remains constant throughout the motion since there is no horizontal acceleration. Therefore, after 2 seconds:
\[ v_x = v_{x0} = 160\sqrt{3} \, \text{ft/s} \]

### Step 3: Calculate the vertical velocity after 2 seconds
The vertical component of velocity changes due to the acceleration caused by gravity. The formula for vertical velocity at time \( t \) is:
\[ v_y = v_{y0} - g t \]
where \( g = 32 \, \text{ft/s}^2 \) (acceleration due to gravity).

Substituting the values:
\[ v_y = 160 - 32 \times 2 = 160 - 64 = 96 \, \text{ft/s} \]

### Step 4: Calculate the resultant velocity
Now that we have both components of the velocity after 2 seconds, we can find the magnitude of the resultant velocity \( v \) using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(160\sqrt{3})^2 + (96)^2} = \sqrt{76800 + 9216} = \sqrt{86016} \]
Calculating this gives:
\[ v \approx 293.3 \, \text{ft/s} \]

### Step 5: Calculate the direction of the velocity
To find the angle \( \theta \) of the velocity with respect to the horizontal, we use the tangent function:
\[ \tan(\theta) = \frac{v_y}{v_x} = \frac{96}{160\sqrt{3}} \]
Thus,
\[ \theta = \tan^{-1}\left(\frac{96}{160\sqrt{3}}\right) \]

### Final Result
The direction of the velocity after 2 seconds is given by the angle \( \theta \).

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