A point mass is projected, making an acute angle with the horizontal. If angle between velocity `vecv` and acceleration `vecg` is `theta`, then `theta` is given by

Options

A. `0^(@) lt theta lt 90^(@)`
B. `theta=90^(@)`
C. `theta=90^(@)`
D. `0^(@) lt theta lt 180^(@)`

Answer

To find the angle \( \theta \) between the velocity vector \( \vec{v} \) and the acceleration vector \( \vec{g} \) of a point mass projected at an acute angle with the horizontal, we can follow these steps:

### Step 1: Understand the Vectors
- The velocity vector \( \vec{v} \) is directed at an angle \( \alpha \) above the horizontal.
- The acceleration vector \( \vec{g} \) (due to gravity) is directed vertically downward.

### Step 2: Identify the Angle \( \theta \)
- The angle \( \theta \) is defined as the angle between the velocity vector \( \vec{v} \) and the acceleration vector \( \vec{g} \).

### Step 3: Analyze the Angles
- Since \( \alpha \) is the angle of projection above the horizontal, the angle between the horizontal and \( \vec{g} \) is \( 90^\circ \).
- Therefore, the angle \( \theta \) can be expressed in terms of \( \alpha \) as:
\[ \theta = 90^\circ + \alpha \]

### Step 4: Determine the Range of \( \theta \)
- Given that \( \alpha \) is an acute angle, it lies in the range \( 0^\circ < \alpha < 90^\circ \).
- Consequently, \( \theta \) will range from:
- Minimum: \( \theta = 90^\circ + 0^\circ = 90^\circ \)
- Maximum: \( \theta = 90^\circ + 90^\circ = 180^\circ \)
- Thus, \( \theta \) will always be in the range \( 90^\circ < \theta < 180^\circ \).

### Step 5: Conclusion
- Therefore, the angle \( \theta \) between the velocity vector \( \vec{v} \) and the acceleration vector \( \vec{g} \) is given by:
\[ 90^\circ < \theta < 180^\circ \]