A body is thrown with the velocity `v_0` at an angle of `theta` to the horizontal. 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)= 9]`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between maximum height and initial velocity
The maximum height \( h \) attained by a projectile is given by the formula:
\[ h = \frac{v_0^2 \sin^2 \theta}{2g} \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 2: Substitute the known values into the height equation
Given that the maximum height \( h = 5 \, \text{m} \):
\[ 5 = \frac{v_0^2 \sin^2 \theta}{2 \times 10} \]
This simplifies to:
\[ 5 = \frac{v_0^2 \sin^2 \theta}{20} \]
Multiplying both sides by 20 gives:
\[ 100 = v_0^2 \sin^2 \theta \]
Thus, we have:
\[ v_0^2 \sin^2 \theta = 100 \quad \text{(Equation 1)} \]

### Step 3: Use the radius of curvature at the highest point
At the highest point of the trajectory, the radius of curvature \( r \) is given by:
\[ \frac{v^2}{g} = r \]
At the highest point, the vertical component of the velocity is zero, and the horizontal component is \( v_0 \cos \theta \). Therefore, we can write:
\[ \frac{(v_0 \cos \theta)^2}{g} = r \]
Substituting \( g = 10 \, \text{m/s}^2 \) and \( r = 3 \, \text{m} \):
\[ \frac{(v_0 \cos \theta)^2}{10} = 3 \]
This simplifies to:
\[ (v_0 \cos \theta)^2 = 30 \quad \text{(Equation 2)} \]

### Step 4: Relate \( \sin^2 \theta \) and \( \cos^2 \theta \)
From the Pythagorean identity, we know:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Let \( \cos^2 \theta = x \). Then \( \sin^2 \theta = 1 - x \).

### Step 5: Substitute \( \sin^2 \theta \) and \( \cos^2 \theta \) into the equations
From Equation 1:
\[ v_0^2 (1 - x) = 100 \quad \text{(Equation 3)} \]
From Equation 2:
\[ v_0^2 x = 30 \quad \text{(Equation 4)} \]

### Step 6: Solve the equations simultaneously
From Equation 4, we can express \( v_0^2 \):
\[ v_0^2 = \frac{30}{x} \]
Substituting this into Equation 3:
\[ \frac{30}{x} (1 - x) = 100 \]
Multiplying through by \( x \) gives:
\[ 30(1 - x) = 100x \]
Expanding and rearranging:
\[ 30 - 30x = 100x \]
\[ 30 = 130x \]
\[ x = \frac{30}{130} = \frac{3}{13} \]

### Step 7: Find \( v_0^2 \)
Substituting \( x \) back into Equation 4:
\[ v_0^2 = \frac{30}{\frac{3}{13}} = 30 \cdot \frac{13}{3} = 130 \]
Thus:
\[ v_0 = \sqrt{130} \approx 11.4 \, \text{m/s} \]

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

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