Math Problem Statement

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

Solution

To solve the equation:

15=25tan(65)219.81v02cos2(65)62515 = \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 v0v_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=25tan(65)219.81625v02cos2(65)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:

    9.81625v02cos2(65)\frac{9.81 \cdot 625}{v_0^2 \cdot \cos^2(65^\circ)}

    Which can be rewritten as:

    6126.25v02cos2(65)\frac{6126.25}{v_0^2 \cdot \cos^2(65^\circ)}

  3. Substitute the trigonometric values:

    • tan(65)2.1445\tan(65^\circ) \approx 2.1445
    • cos(65)0.4226\cos(65^\circ) \approx 0.4226
    • cos2(65)0.1786\cos^2(65^\circ) \approx 0.1786
  4. Calculate each part:

    Substitute the values:

    15=252.1445216126.25v020.178615 = \frac{25 \cdot 2.1445 - 21}{\frac{6126.25}{v_0^2 \cdot 0.1786}}

    Simplify:

    15=53.6125216126.25v020.178615 = \frac{53.6125 - 21}{\frac{6126.25}{v_0^2 \cdot 0.1786}}

    Which simplifies further to:

    15=32.61256126.25v020.178615 = \frac{32.6125}{\frac{6126.25}{v_0^2 \cdot 0.1786}}

  5. Invert and solve for v02v_0^2:

    Invert the fraction:

    15=32.6125×v020.17866126.2515 = 32.6125 \times \frac{v_0^2 \cdot 0.1786}{6126.25}

    Solve for v02v_0^2:

    156126.25=32.6125v020.178615 \cdot 6126.25 = 32.6125 \cdot v_0^2 \cdot 0.1786

    91893.75=5.825v0291893.75 = 5.825 \cdot v_0^2

    v02=91893.755.825v_0^2 = \frac{91893.75}{5.825}

    v0215772.18v_0^2 \approx 15772.18

    v015772.18125.59v_0 \approx \sqrt{15772.18} \approx 125.59

  6. Final Answer:

    The initial velocity v0v_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.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Projectile Motion
Algebra

Formulas

v_0 = \sqrt{\frac{91893.75}{5.825}}
15 = \frac{25 \cdot \tan(65^\circ) - 21}{\frac{9.81}{v_0^2 \cdot \cos^2(65^\circ)} \cdot 625}
\tan(65^\circ) \approx 2.1445
\cos(65^\circ) \approx 0.4226
\cos^2(65^\circ) \approx 0.1786

Theorems

Trigonometric Identities
Laws of Projectile Motion

Suitable Grade Level

Grades 11-12