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To answer your questions:

  1. The initial velocity of the cannonball is already provided in the question as 85.76 m/s. Therefore, the initial velocity is 85.76 m/s.

  2. To find the maximum height reached by the cannonball, we can use the equations of projectile motion. In this case, we need to consider the vertical motion of the cannonball. The maximum height is achieved when the vertical component of the velocity becomes zero.

The initial velocity of the cannonball can be divided into horizontal and vertical components. The vertical component can be found using the initial velocity and the angle of elevation.

Vertical component of the initial velocity (v_y) = initial velocity * sin(angle) v_y = 85.76 m/s * sin(45°) = 85.76 m/s * 0.7071 ≈ 60.66 m/s

Now, we can find the time it takes for the vertical component of velocity to become zero using the equation:

v_y = initial vertical velocity + (acceleration * time)

At the maximum height, v_y becomes zero, and the acceleration acting on the object is the acceleration due to gravity (-9.8 m/s²). Solving the equation for time:

0 = 60.66 m/s + (-9.8 m/s²) * time

Solving for time: 60.66 m/s = 9.8 m/s² * time time = 60.66 m/s / 9.8 m/s² ≈ 6.19 s

Now, we can find the maximum height (h) reached using the equation:

h = initial vertical velocity * time + (1/2) * acceleration * time²

h = 60.66 m/s * 6.19 s + (1/2) * (-9.8 m/s²) * (6.19 s)² h ≈ 188.84 m

Therefore, the maximum height reached by the cannonball is approximately 188.84 meters.

  1. The initial horizontal velocity can be found using the initial velocity and the angle of elevation.

Horizontal component of the initial velocity (v_x) = initial velocity * cos(angle) v_x = 85.76 m/s * cos(45°) = 85.76 m/s * 0.7071 ≈ 60.66 m/s

Therefore, the initial horizontal velocity of the cannonball is approximately 60.66 m/s.

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