The highest barrier that a projectile can clear is 13.4 m, when the projectile is launched at

an angle of 11.0o above the horizontal. What is the projectile’s launch speed?

As you know, the max height reached is

h = (v sinθ)^2/2g

To find the projectile's launch speed, we can use the equations of projectile motion. The given information is that the highest barrier the projectile can clear is 13.4 m and the launch angle is 11.0o above the horizontal.

The key equation we will use is the vertical distance formula:

d = (v^2 * sin^2θ) / (2 * g)

Where:
d = vertical distance traveled by the projectile (13.4 m)
v = launch speed of the projectile (unknown)
θ = launch angle (11.0o)
g = acceleration due to gravity (9.8 m/s^2)

Now, we can rearrange the formula to solve for v:

v^2 = (d * 2 * g) / sin^2θ

Let's substitute the given values into the formula:

v^2 = (13.4 m * 2 * 9.8 m/s^2) / sin^2(11.0o)

Now, we can calculate v:

v^2 = (26.8 m^2/s^2 * 9.8 m/s^2) / sin^2(11.0o)

v^2 = 262.64 m^2/s^2 / (0.03143)^2

v^2 = 262.64 m^2/s^2 / 0.0009877

v^2 = 265833.43

Taking the square root of both sides, we find:

v ≈ 515.61 m/s

Therefore, the projectile's launch speed is approximately 515.61 m/s.