The angle of which the velocity vector of a projectile thrown with a velocity u at an angle `theta` to the horizontal will take 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)/(vsin theta-"gt"))`
D. `tan^(-1)((vsin theta-"gt")/(v cos theta))`

Explanation

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

### Step-by-Step Solution:

1. **Identify Initial Velocity Components**:
The initial velocity \( u \) is given at an angle \( \theta \) to the horizontal. We can resolve this velocity into horizontal and vertical components:
- Horizontal component \( u_x = u \cos \theta \)
- Vertical component \( u_y = u \sin \theta \)

2. **Determine the Vertical Velocity After Time \( t \)**:
The vertical velocity \( v_y \) after time \( t \) can be calculated using the equation of motion:
\[ v_y = u_y - g t \]
where \( g \) is the acceleration due to gravity. Substituting \( u_y \):
\[ v_y = u \sin \theta - g t \]

3. **Determine the Horizontal Velocity After Time \( t \)**:
The horizontal velocity \( v_x \) remains constant throughout the projectile's motion (since there is no horizontal acceleration):
\[ v_x = u_x = u \cos \theta \]

4. **Calculate the Angle with the Horizontal**:
The angle \( \phi \) that the velocity vector makes with the horizontal can be found using the tangent function:
\[ \tan \phi = \frac{v_y}{v_x} \]
Substituting the values of \( v_y \) and \( v_x \):
\[ \tan \phi = \frac{u \sin \theta - g t}{u \cos \theta} \]

5. **Find the Angle \( \phi \)**:
To find the angle \( \phi \), we take the arctangent:
\[ \phi = \tan^{-1} \left( \frac{u \sin \theta - g t}{u \cos \theta} \right) \]

### Final Answer:
The angle \( \phi \) that the velocity vector makes with the horizontal after time \( t \) is:
\[ \phi = \tan^{-1} \left( \frac{u \sin \theta - g t}{u \cos \theta} \right) \]