A cannon ball is fired with a velocity of 200 m\s2 at an < of 25° above ground. The total flight time is 45.6 seconds. (Neglect air resistance.)

a. Calculate the initial horizontal\vertical components of the velocity
b. Calculate maximum height of the cannon ball above ground

a. To calculate the initial horizontal and vertical components of the velocity, we can use the following formulas:

Horizontal component of velocity (Vx) = Initial velocity (V) * Cos(angle)
Vertical component of velocity (Vy) = Initial velocity (V) * Sin(angle)

Given:
Initial velocity (V) = 200 m/s
Angle (θ) = 25°

Plugging the values into the formulas, we get:

Vx = 200 m/s * Cos(25°)
Vy = 200 m/s * Sin(25°)

Calculating the values:

Vx ≈ 200 m/s * 0.9063 ≈ 181.26 m/s (rounded to two decimal places)
Vy ≈ 200 m/s * 0.4239 ≈ 84.78 m/s (rounded to two decimal places)

Therefore, the initial horizontal component of velocity (Vx) is approximately 181.26 m/s, and the initial vertical component of velocity (Vy) is approximately 84.78 m/s.

b. To calculate the maximum height of the cannonball above the ground, we can use the following equation:

Maximum height (H) = (Vy^2) / (2 * g)

Given:
Vertical component of velocity (Vy) = 84.78 m/s
Acceleration due to gravity (g) = 9.8 m/s^2 (assuming it is on Earth)

Plugging the values into the equation, we get:

H = (84.78 m/s)^2 / (2 * 9.8 m/s^2)

Calculating the value:

H ≈ (7201.6484 m^2/s^2) / (19.6 m/s^2) ≈ 368.13 m (rounded to two decimal places)

Therefore, the maximum height of the cannonball above the ground is approximately 368.13 meters.