15=25⋅tan(65∘)−21​(9.81)⋅v02​cos2(65∘)625​


To solve the equation:

\[
15 = \frac{25 \cdot \tan(65^\circ) - 21}{\frac{9.81}{v_0^2 \cos^2(65^\circ)} \cdot 625}
\]

we need to first simplify and solve for the unknown, which appears to be \( v_0 \) (initial velocity). Let's go through the steps to solve this equation.

### Step-by-Step Solution:

1. **Simplify the expression**:
   
   Rewrite the equation more clearly:

   \[
   15 = \frac{25 \cdot \tan(65^\circ) - 21}{\frac{9.81 \cdot 625}{v_0^2 \cdot \cos^2(65^\circ)}}
   \]

2. **Simplify the denominator**:

   The denominator is:

   \[
   \frac{9.81 \cdot 625}{v_0^2 \cdot \cos^2(65^\circ)}
   \]

   Which can be rewritten as:

   \[
   \frac{6126.25}{v_0^2 \cdot \cos^2(65^\circ)}
   \]

3. **Substitute the trigonometric values**:

   - \(\tan(65^\circ) \approx 2.1445\)
   - \(\cos(65^\circ) \approx 0.4226\)
   - \(\cos^2(65^\circ) \approx 0.1786\)

4. **Calculate each part**:

   Substitute the values:

   \[
   15 = \frac{25 \cdot 2.1445 - 21}{\frac{6126.25}{v_0^2 \cdot 0.1786}}
   \]

   Simplify:

   \[
   15 = \frac{53.6125 - 21}{\frac{6126.25}{v_0^2 \cdot 0.1786}}
   \]

   Which simplifies further to:

   \[
   15 = \frac{32.6125}{\frac{6126.25}{v_0^2 \cdot 0.1786}}
   \]

5. **Invert and solve for \(v_0^2\)**:

   Invert the fraction:

   \[
   15 = 32.6125 \times \frac{v_0^2 \cdot 0.1786}{6126.25}
   \]

   Solve for \(v_0^2\):

   \[
   15 \cdot 6126.25 = 32.6125 \cdot v_0^2 \cdot 0.1786
   \]

   \[
   91893.75 = 5.825 \cdot v_0^2
   \]

   \[
   v_0^2 = \frac{91893.75}{5.825}
   \]

   \[
   v_0^2 \approx 15772.18
   \]

   \[
   v_0 \approx \sqrt{15772.18} \approx 125.59
   \]

6. **Final Answer**:

   The initial velocity \( v_0 \) is approximately **125.59 m/s**.

Would you like more details on any part of this solution or have any specific questions?

### Further Related Questions:
1. How does the tangent function relate to the angle in this context?
2. What is the significance of using trigonometric identities in solving physics problems?
3. How can you use this formula in real-life projectile motion problems?
4. What other forces could affect the velocity of a projectile in a similar problem?
5. Can this approach be used to solve problems with air resistance?

**Tip:** In problems involving trigonometric functions, always consider simplifying by using trigonometric identities or standard values for specific angles.

15 = 25⋅tan(65∘)−21​(9.81)⋅v02​cos2(65∘)625​

Learn how to solve a complex projectile motion problem involving trigonometric functions like tan(65°) and cos(65°) to find the initial velocity (v0). This problem demonstrates key physics and algebraic concepts, suitable for students in Grades 11-12.

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