A cannon tilted up at a 34.0^\circ angle fires a cannon ball at 69.0 m/s from atop a 23.0 m-high fortress wall. What is the ball's impact speed on the ground below?

total energy at launch = KE at bottom

1/2 m 69^2 + mg(23)= 1/2 m vfinal^2
solve for vfinal

To find the impact speed of the cannonball on the ground below, we can break down the problem into two parts: the horizontal and vertical components of the cannonball's motion.

First, let's determine the initial velocity of the cannonball in the horizontal direction. Since there is no horizontal acceleration, the initial horizontal velocity will remain constant throughout the motion. We can use trigonometry to find this component.

The horizontal component of the initial velocity can be calculated using the formula:

Vx = V * cos(theta)

where Vx is the horizontal component of the velocity, V is the initial velocity of the cannonball, and theta is the angle of inclination (34.0 degrees in this case).

Substituting the given values into the equation, we get:

Vx = 69.0 m/s * cos(34.0 degrees)
Vx ≈ 57.30 m/s

Next, let's determine the initial vertical velocity of the cannonball. The vertical component of the initial velocity can be calculated using the formula:

Vy = V * sin(theta)

where Vy is the vertical component of the velocity, V is the initial velocity of the cannonball, and theta is the angle of inclination (34.0 degrees).

Substituting the given values into the equation, we get:

Vy = 69.0 m/s * sin(34.0 degrees)
Vy ≈ 37.56 m/s

Now, we can analyze the vertical motion of the cannonball. The vertical motion can be treated as a projectile under the influence of gravity. We assume that the acceleration due to gravity is approximately 9.8 m/s^2 downward.

Using the kinematic equation:

h = Vy * t - (1/2) * g * t^2

where h is the initial height (23.0 m in this case), Vy is the vertical component of the velocity, g is the acceleration due to gravity, and t is the time of flight.

Since we are interested in the time it takes for the cannonball to hit the ground, we set h to zero and solve for t:

0 = (37.56 m/s) * t - (0.5 * 9.8 m/s^2) * t^2

Rearranging the equation, we get:

4.9 * t^2 - 37.56 * t = 0

Factoring out t, we have:

t * (4.9 * t - 37.56) = 0

Solving each term separately:

t = 0 (ignoring this solution since it represents the time at the start)
4.9 * t - 37.56 = 0

Solving for t:

4.9 * t = 37.56

t = 37.56 / 4.9
t ≈ 7.67 seconds

Now, we can use the time of flight to find the horizontal displacement of the cannonball using the horizontal component of the velocity:

dx = Vx * t

Substituting the known values:

dx = 57.30 m/s * 7.67 s
dx ≈ 439.94 m

Finally, we can find the impact speed of the cannonball on the ground below by using the horizontal and vertical displacements:

Impact speed = √(dx^2 + h^2)

Substituting the known values:

Impact speed = √(439.94^2 + 23.0^2)
Impact speed ≈ 444.41 m/s

Therefore, the ball's impact speed on the ground below is approximately 444.41 m/s.