A body is thrown with the velocity `v_0` at an angle of `theta` to the horizon. Determine `v_0 "in" m s^-1` if the maximum height attained by the body is `5 m` and at the highest point of its trajectory the radius of curvature is `r = 3 m`. Neglect air resistance. `[Use sqrt(80) as 9]`.

Explanation

To solve the problem, we need to determine the initial velocity \( v_0 \) of a projectile thrown at an angle \( \theta \) given the maximum height \( h = 5 \, m \) and the radius of curvature \( r = 3 \, m \) at the highest point of its trajectory. We will neglect air resistance.

### Step-by-Step Solution:

1. **Understanding the Radius of Curvature**:
The radius of curvature \( r \) at the highest point of the projectile's trajectory can be expressed as:
\[ r = \frac{v_0^2 \cos^2 \theta}{g} \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, m/s^2 \)).

2. **Understanding Maximum Height**:
The maximum height \( h \) attained by the projectile can be expressed as:
\[ h = \frac{v_0^2 \sin^2 \theta}{2g} \]

3. **Relating the Two Equations**:
From the two equations, we can express \( v_0^2 \) in terms of \( r \) and \( h \):
- From the radius of curvature equation:
\[ v_0^2 = rg \]
- From the maximum height equation:
\[ v_0^2 = 2hg \cdot \frac{1}{\sin^2 \theta} \]

4. **Adding the Two Expressions**:
We can add the two expressions for \( v_0^2 \):
\[ rg + 2hg = v_0^2 (\sin^2 \theta + \cos^2 \theta) \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we have:
\[ v_0^2 = rg + 2hg \]

5. **Substituting Values**:
Now, substituting the known values:
- \( r = 3 \, m \)
- \( h = 5 \, m \)
- \( g = 10 \, m/s^2 \)
\[ v_0^2 = (3)(10) + 2(5)(10) = 30 + 100 = 130 \]

6. **Calculating \( v_0 \)**:
Taking the square root to find \( v_0 \):
\[ v_0 = \sqrt{130} \]
Since \( \sqrt{80} \approx 9 \), we can express \( \sqrt{130} \) as:
\[ \sqrt{130} = \sqrt{(80 + 50)} \approx \sqrt{80} + \sqrt{50} \approx 9 + 7 \approx 16 \]
However, for exact calculations:
\[ v_0 = \sqrt{130} \approx 11.4 \, m/s \]

### Final Answer:
The initial velocity \( v_0 \) is approximately \( 11.4 \, m/s \).