During a training exercise a cannon ball is fired with a muzzle velocity of

84.9 m/s at an angle of 45o above the horizontal. Determine how far from the cannon the ball will land when it hits the ground.

R = Vo^2*sin(2A) / g.

R = (84.9)^2*sin90 / 9.8 = 736 m.

To determine how far the cannonball will land from the cannon, we need to analyze the projectile motion of the cannonball.

First, let's break the initial velocity of the cannonball into its horizontal and vertical components. The horizontal component of the velocity remains constant throughout the projectile motion, while the vertical component changes due to the influence of gravity.

Given that the muzzle velocity of the cannonball is 84.9 m/s and the launch angle is 45 degrees above the horizontal, we can calculate the initial horizontal and vertical velocities.

Horizontal velocity (Vx) = initial velocity * cos(angle)
Vx = 84.9 m/s * cos(45 degrees)
Vx = 84.9 m/s * √2 / 2
Vx ≈ 60 m/s

Vertical velocity (Vy) = initial velocity * sin(angle)
Vy = 84.9 m/s * sin(45 degrees)
Vy = 84.9 m/s * √2 / 2
Vy ≈ 60 m/s

Now, we can determine the time of flight for the cannonball. The time of flight is the total time it takes for the cannonball to reach its maximum height and fall back down to the ground.

Since the vertical motion of the cannonball follows a standard free-fall motion, we can use the following formula:

Time of flight = 2 * (Vertical velocity) / acceleration due to gravity

The acceleration due to gravity (g) is approximately 9.8 m/s^2.

Time of flight = 2 * 60 m/s / 9.8 m/s^2
Time of flight ≈ 12.24 s

Now, we can calculate the horizontal distance traveled by the cannonball using the horizontal velocity (Vx) and the time of flight.

Horizontal distance = Horizontal velocity * Time of flight
Horizontal distance = 60 m/s * 12.24 s
Horizontal distance ≈ 734.4 m

Therefore, the cannonball will land approximately 734.4 meters away from the cannon.