2) A ball is hit at an angle of 36 degrees to the horizontal from 14 feet above the ground. If it hits the ground 26 feet away (in horizontal distance), what was the initial velocity? Assume there is no air resistance.

To solve this problem, we need to use the principle of projectile motion. Let's break down the given information:

1. The ball is hit at an angle of 36 degrees to the horizontal.
2. The ball is launched from a height of 14 feet above the ground.
3. The ball hits the ground 26 feet away (horizontally).

We are required to find the initial velocity of the ball. To do this, we can use the following equations of motion:

1. Horizontal distance (R): R = V₀ * t * cos(theta)
2. Vertical distance (H): H = V₀ * t * sin(theta) - (1/2) * g * t^2
3. Time of flight (T): T = 2 * (V₀ * sin(theta)) / g

Here, V₀ represents the initial velocity, theta is the launch angle (36 degrees), and g is the acceleration due to gravity (approximately 32.2 ft/s^2).

Let's solve step by step:

1. First, we need to find the time of flight using the equation T = 2 * (V₀ * sin(theta)) / g.
Plugging in the given values, we get: T = 2 * (V₀ * sin(36°)) / 32.2.

2. Next, we can determine the horizontal distance using the equation R = V₀ * t * cos(theta).
Plugging in the given values for R and time of flight, we get: 26 ft = V₀ * T * cos(36°).

3. Now, we can substitute the value of T from step 1 into the equation from step 2 and solve for V₀.
Using the given values, we get: 26 ft = V₀ * (2 * (V₀ * sin(36°)) / 32.2) * cos(36°).

By solving this equation, you can find the initial velocity (V₀) of the ball.