The highest barrier that a projectile can clear is 13.8 m, when the projectile is launched at an angle of 14.9° above the horizontal. What is the projectile's launch speed?

v^2=2(9.8)(13.8)/sin(14.9)

=32.43

To find the projectile's launch speed, we can use the concepts of projectile motion and trigonometry.

First, let's break down the given information:
- The highest barrier cleared by the projectile is 13.8 m.
- The angle of launch above the horizontal is 14.9°.

When analyzing projectile motion, we can separate the motion into horizontal and vertical components.

1. Vertical Component:
The maximum height reached by the projectile occurs when its vertical velocity becomes zero. We can use the kinematic equation for vertical displacement to find the initial vertical velocity (viy) of the projectile.

Using the equation:
viy² = v₀y² - 2 * g * Δy
where viy is the initial vertical velocity, v₀y is the vertical component of the initial velocity, g is the acceleration due to gravity (approximately -9.8 m/s²), and Δy is the vertical displacement (13.8 m).

Since the projectile is launched at an angle, we can use the relationship between the initial velocity and its vertical and horizontal components (v₀ = √(v₀x² + v₀y²)).

2. Horizontal Component:
The horizontal motion of the projectile is governed by a constant velocity. The horizontal component of the initial velocity (v₀x) can be found using the equation v₀x = v₀ * cos(θ), where v₀ is the initial velocity of the projectile and θ is the launch angle.

3. Launch Speed:
The launch speed (v₀) is the magnitude of the initial velocity (v₀ = √(v₀x² + v₀y²)).

Now let's calculate the launch speed:

Step 1: Calculate the initial vertical velocity (viy):
viy² = v₀y² - 2 * g * Δy
0² = v₀ * sin(θ)² - 2 * (-9.8) * 13.8
-2 * (-9.8) * 13.8 = v₀ * sin(θ)²
v₀ * sin(θ)² = 270.612

Step 2: Calculate the initial horizontal velocity (v₀x):
v₀x = v₀ * cos(θ)

Step 3: Calculate the magnitude of the initial velocity (v₀):
v₀ = √(v₀x² + v₀y²)
v₀ = √((v₀ * cos(θ))² + 270.612²)

To solve this equation, we'll substitute v₀ with a variable "x", so the equation becomes:
x = √((x * cos(θ))² + 270.612²)

This equation doesn't have a straightforward algebraic solution. We can solve it numerically using numerical methods such as the Newton-Raphson method or by using graphing software.

Using the Newton-Raphson method:
1. Guess an initial value for "x."
2. Calculate the function value and its derivative using that guess.
3. Update the guess using the formula: x - (f(x) / f'(x)).
4. Repeat steps 2-3 until the guess converges to a solution.

Alternatively, you can input the equation into graphing software and find the x-value where the equation equals zero.

Once you find the value for "x," that will correspond to the launch speed (v₀) of the projectile.