We are tasked with finding the initial velocity $V_0$ in this projectile motion problem.

The given equation is:

$1.33 \, \text{m} = 8.60 \, \text{m} \cdot \tan(20^\circ) - \frac{4.905 \, \text{m/s}^2 \cdot (8.60 \, \text{m})^2}{V_0^2 \cdot \cos^2(20^\circ)}$

Step-by-Step Solution:

1. First, simplify the known trigonometric values:

$\tan(20^\circ) \approx 0.3640 \quad \text{and} \quad \cos(20^\circ) \approx 0.9397$

2. Substitute the known values:

$1.33 = 8.60 \cdot 0.3640 - \frac{4.905 \cdot (8.60)^2}{V_0^2 \cdot (0.9397)^2}$

3. Simplify the terms:

$1.33 = 3.1304 - \frac{4.905 \cdot 73.96}{V_0^2 \cdot 0.8830}$ $1.33 = 3.1304 - \frac{362.14}{0.8830 \cdot V_0^2}$ $1.33 = 3.1304 - \frac{410.2}{V_0^2}$

4. Move the constant terms to the other side:

$1.33 - 3.1304 = - \frac{410.2}{V_0^2}$ $-1.8004 = - \frac{410.2}{V_0^2}$ $1.8004 = \frac{410.2}{V_0^2}$

5. Solve for $V_0^2$:

$V_0^2 = \frac{410.2}{1.8004}$ $V_0^2 \approx 227.84$

6. Finally, take the square root to find $V_0$:

$V_0 \approx \sqrt{227.84} \approx 15.1 \, \text{m/s}$

Final Answer:

The initial velocity $V_0$ is approximately 15.1 m/s.

Would you like more details on any step or further explanation?

Related Questions:

1. How does the angle of projection affect the range of a projectile?
2. How is the horizontal range of a projectile derived?
3. What happens if the angle is increased in this scenario?
4. How would air resistance change the calculations for projectile motion?
5. What is the difference between horizontal and vertical components of velocity?

Tip:

Always break down equations step by step, especially in projectile motion problems, to clearly identify the horizontal and vertical components!