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Teniendo en cuenta la situación, aplica identidades trigonométricas, para reescribir la fórmula de ym de otro modo. Usa las 2 fórmulas (la dada inicialmente y la obtenida en el punto 1), y calcula la velocidad con la que debe ser lanzado un objeto, para que alcance una altura máxima h, cuando se lanza con una inclinación de 30°.

h= 168 en metros. Compara los resultados obtenidos con cada una de las fórmulas en el punto anterior, y justifica por qué esos resultados son iguales o diferentes.

To tackle this problem, we will first re-examine the given formula and apply trigonometric identities to rewrite it as required. Then, we will calculate the initial velocity needed to reach the specified height using both the original and transformed formulas, followed by a comparison of the results.

### Step 1: Analyze and Reformat the Formula
The provided formula for maximum height \( y_m \) is:

\[
y_m = \frac{V_0^2 \sin^2(\theta)}{2g}
\]

where:
- \( \theta \) is the angle of inclination,
- \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)),
- \( V_0 \) is the initial velocity.

**Trigonometrical identity:**

We know from trigonometry that:

\[
\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}
\]

Using this identity, we can rewrite the formula:

\[
y_m = \frac{V_0^2 \cdot \frac{1 - \cos(2\theta)}{2}}{2g} = \frac{V_0^2 (1 - \cos(2\theta))}{4g}
\]

### Step 2: Calculate Initial Velocity for Given Maximum Height
Given that the height \( h = 168 \, \text{meters} \) and \( \theta = 30^\circ \), we will calculate the required velocity \( V_0 \) using both the original and the transformed formulas.

#### Using the Original Formula:

\[
168 = \frac{V_0^2 \sin^2(30^\circ)}{2 \times 9.8}
\]

Since \( \sin(30^\circ) = \frac{1}{2} \):

\[
168 = \frac{V_0^2 \cdot \left(\frac{1}{2}\right)^2}{2 \times 9.8} = \frac{V_0^2 \cdot \frac{1}{4}}{19.6}
\]

\[
168 \times 19.6 = \frac{V_0^2}{4}
\]

\[
V_0^2 = 168 \times 19.6 \times 4 = 13171.2
\]

\[
V_0 = \sqrt{13171.2} \approx 114.75 \, \text{m/s}
\]

#### Using the Transformed Formula:

\[
168 = \frac{V_0^2 (1 - \cos(60^\circ))}{4 \times 9.8}
\]

Since \( \cos(60^\circ) = \frac{1}{2} \):

\[
168 = \frac{V_0^2 \cdot \left(1 - \frac{1}{2}\right)}{4 \times 9.8} = \frac{V_0^2 \cdot \frac{1}{2}}{39.2}
\]

\[
168 \times 39.2 = \frac{V_0^2}{2}
\]

\[
V_0^2 = 168 \times 39.2 \times 2 = 13171.2
\]

\[
V_0 = \sqrt{13171.2} \approx 114.75 \, \text{m/s}
\]

### Step 3: Compare the Results
The results from both formulas yield the same initial velocity:

\[ V_0 \approx 114.75 \, \text{m/s} \]

### Justification:
The two formulas are algebraically equivalent. The application of trigonometric identities does not change the outcome, but rather presents the formula in a different form. Since both representations ultimately describe the same physical reality, they yield the same numerical result.

### Questions for Further Exploration:
1. What would happen to the required velocity if the angle of inclination is doubled?
2. How does the maximum height change if the velocity is increased by 10%?
3. Can you derive a similar formula for the maximum horizontal range of a projectile?
4. How would air resistance alter the result obtained using these formulas?
5. What are the real-world applications of these equations in engineering?

### Tip:
Always check the trigonometric identities you use in physics problems to ensure they fit the context of the equation. This can simplify the solution or provide insights into the problem.

Learn how to calculate the initial velocity of a projectile to reach a maximum height of 168 meters using trigonometric identities. Compare results derived from the original and transformed formulas. Suitable for high school students studying trigonometry and projectile motion.

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