The maximum angle to the horizontal at which a stone can be thrown so that it always moves away from the thrower will be :

Answer

To solve the problem of finding the maximum angle to the horizontal at which a stone can be thrown so that it always moves away from the thrower, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to determine the maximum angle \( \theta \) at which a stone can be thrown such that it always moves away from the thrower. This means that the radial distance from the thrower to the stone must always increase.

2. **Defining the Motion**:
When the stone is thrown with an initial velocity \( V \) at an angle \( \theta \), it has two components of velocity:
- Horizontal component: \( V_x = V \cos \theta \)
- Vertical component: \( V_y = V \sin \theta \)

3. **Position Vector**:
The position vector \( \mathbf{r} \) of the stone at time \( t \) can be expressed as:
\[ \mathbf{r} = (V \cos \theta) t \hat{i} + \left(V \sin \theta t - \frac{1}{2} g t^2\right) \hat{j} \]

4. **Velocity Vector**:
The velocity vector \( \mathbf{v} \) at time \( t \) is given by:
\[ \mathbf{v} = V \cos \theta \hat{i} + \left(V \sin \theta - g t\right) \hat{j} \]

5. **Condition for Moving Away**:
For the stone to always move away from the thrower, the dot product of the position vector \( \mathbf{r} \) and the velocity vector \( \mathbf{v} \) must be positive:
\[ \mathbf{r} \cdot \mathbf{v} > 0 \]

6. **Calculating the Dot Product**:
\[ \mathbf{r} \cdot \mathbf{v} = \left(V \cos \theta t\right)(V \cos \theta) + \left(V \sin \theta t - \frac{1}{2} g t^2\right)\left(V \sin \theta - g t\right) \]
Expanding this gives us:
\[ = V^2 \cos^2 \theta t + \left(V \sin \theta t - \frac{1}{2} g t^2\right)(V \sin \theta - g t) \]

7. **Simplifying the Expression**:
After simplification, we need to ensure that the resulting expression is always greater than zero. This leads to a quadratic inequality in \( t \).

8. **Finding Conditions on \( \theta \)**:
For the quadratic expression to be always positive, we analyze the coefficients. The discriminant of the quadratic must be less than zero:
\[ 9/8 V^2 \sin^2 \theta < V^2 \]

9. **Solving for \( \theta \)**:
Rearranging gives:
\[ \sin^2 \theta < \frac{8}{9} \implies \sin \theta < \frac{2\sqrt{2}}{3} \]
Thus, the maximum angle \( \theta \) can be found using:
\[ \theta < \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \]

10. **Final Answer**:
The maximum angle \( \theta \) at which the stone can be thrown so that it always moves away from the thrower is:
\[ \theta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \]