A missile launches at a velocity of 30.0 m/s at an angle of 30.0* to the normal. What is the maximum height the missile attains?

(((30)*cos(30))^2)/(2*9.8) )

Nevermind...

v2-u2=2as
Here a = -9.8 m/s 2
Putting in the values
0 - 225 = 2*(-9.8)*S
S = 11.5 m

That's it right>

To calculate the maximum height attained by the missile, we can use the kinematic equations of projectile motion. The formula you provided seems to be a calculation based on the range of the projectile, rather than the maximum height. Let's first review the correct approach.

In projectile motion, the initial velocity can be divided into horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component is influenced by gravity.

Given:
Initial velocity (v₀) = 30.0 m/s
Launch angle (θ) = 30.0°

To find the maximum height (h) reached by the missile, we need to determine the time it takes for the projectile to reach the highest point. This can be done using the vertical component of the initial velocity.

Vertical initial velocity (v₀ₓ) = v₀ * sin(θ)
v₀ₓ = 30.0 m/s * sin(30.0°)
v₀ₓ = 15.0 m/s

Now, we can use this vertical initial velocity to determine the time taken to reach the highest point. At the highest point, the vertical velocity becomes zero.

Using the equation:
v = v₀ + at
0 = v₀ₓ + (-9.8 m/s²) * t_max

We can solve for t_max:
t_max = v₀ₓ / 9.8 m/s²
t_max = 15.0 m/s / 9.8 m/s²
t_max ≈ 1.53 seconds

Now, to find the maximum height:

Using the equation:
d = v₀ * t + (1/2) * a * t²

Here, d is the displacement in the vertical direction.

Plugging in the values:
h = v₀ₓ * t_max + (1/2) * (-9.8 m/s²) * t_max²
h = 15.0 m/s * 1.53 s + (1/2) * (-9.8 m/s²) * (1.53 s)²

Calculating:
h ≈ 22.7 meters

Therefore, the maximum height attained by the missile is approximately 22.7 meters.