The angle which the velocity vector of a projectile thrown with a velocity v at an. angle `theta` to the horizontal Will make with the horizontal after time t of its being thrown up is

Options

A. `theta`
B. `tan^(-1) (theta//t)`
C. `tan^(-1) ((v cos theta)/(v sin theta-g t ))`
D. `tan^(-1)((v sin theta - g t)/(v cos theta))`

Answer

To find the angle which the velocity vector of a projectile makes with the horizontal after a time \( t \), we can follow these steps:

### Step 1: Understand the Initial Conditions
A projectile is thrown with an initial velocity \( v \) at an angle \( \theta \) to the horizontal. The initial velocity can be resolved into two components:
- Horizontal component: \( v_x = v \cos \theta \)
- Vertical component: \( v_y = v \sin \theta \)

### Step 2: Analyze the Motion After Time \( t \)
After time \( t \), the horizontal component of the velocity remains unchanged because there is no horizontal acceleration (assuming no air resistance):
- Horizontal velocity at time \( t \): \( v_x = v \cos \theta \)

The vertical component of the velocity changes due to the acceleration due to gravity \( g \):
- Vertical velocity at time \( t \):
\[ v_y = v \sin \theta - g t \]

### Step 3: Determine the Resultant Velocity
The velocity vector after time \( t \) can be represented as:
- \( v_x = v \cos \theta \)
- \( v_y = v \sin \theta - g t \)

### Step 4: Calculate the Angle with the Horizontal
The angle \( \alpha \) that the velocity vector makes with the horizontal can be found using the tangent function:
\[ \tan \alpha = \frac{v_y}{v_x} = \frac{v \sin \theta - g t}{v \cos \theta} \]

### Step 5: Solve for \( \alpha \)
To find the angle \( \alpha \), we take the arctangent of both sides:
\[ \alpha = \tan^{-1}\left(\frac{v \sin \theta - g t}{v \cos \theta}\right) \]

### Final Result
Thus, the angle \( \alpha \) that the velocity vector makes with the horizontal after time \( t \) is given by:
\[ \alpha = \tan^{-1}\left(\frac{v \sin \theta - g t}{v \cos \theta}\right) \]