A cannon rests on the ground at 40 degree angle it fires a ball at 200 m/s how high is the top of trajectory? At what impact speed does it rest? How far away does it land?

To find the answers to these questions, we need to understand the physics of projectile motion and use relevant equations. Let's break it down step by step:

1. How high is the top of the trajectory?
To determine the maximum height of the projectile, we need to find the vertical component of its initial velocity. We can use the following equation:

Vf² = Vi² + 2ad

where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and d is the displacement.

The initial vertical velocity (Vy) can be found using the equation:

Vy = Vi * sin(θ)

where θ is the launch angle.

In this case, Vi = 200 m/s and θ = 40 degrees. Substituting the values, we have:

Vy = 200 m/s * sin(40°)

Now, we can use the equation for vertical motion to find the maximum height (h) the projectile reaches. Assuming the projectile starts and ends at the same vertical position, the initial and final vertical velocities are equal, but opposite in direction. Thus, Vf = -Vy. Using the equation:

Vf² = Vi² + 2ad

we can rewrite it as:

(-Vy)² = Vy² + 2 * (-9.8 m/s²) * h

where g is the acceleration due to gravity, approximately -9.8 m/s². Rearranging the equation, we get:

0 = 2Vy² - 19.6h

Now, substitute the value of Vy we found earlier, and solve for h:

0 = 2 * (200 m/s * sin(40°))² - 19.6h

Simplifying the equation:

0 = 80000sin²(40°) - 19.6h

Now, solve for h using the value of sin²(40°) and evaluate the equation:

h ≈ 403.2 meters

Therefore, the top of the trajectory is approximately 403.2 meters high.

2. Impact speed when the cannonball hits the ground (final velocity):
To find the final velocity when the cannonball lands, we can use the equation for horizontal motion:

Vx = Vi * cos(θ)

where Vx is the horizontal component of the velocity and Vi is the initial velocity.

In this case, Vi = 200 m/s and θ = 40 degrees. Substituting the values, we have:

Vx = 200 m/s * cos(40°)

Evaluate this equation to find the horizontal component of velocity.

3. How far away does it land?
The horizontal displacement (range) of the projectile can be determined using the time of flight (time taken to reach the ground) and the horizontal component of the velocity. The formula for the time of flight is:

t = (2 * Vy) / g

where Vy is the vertical component of the velocity and g is the acceleration due to gravity.

Using the known values, substitute the value of Vy we found earlier, and solve for t:

t = (2 * (200 m/s * sin(40°))) / 9.8 m/s²

Evaluate this equation to find the time of flight.

Once you have the time of flight, you can find the horizontal displacement (range) using the equation:

Range = Vx * t

Substitute the values for Vx and t, which we found earlier, and evaluate the equation to find the range.

By following these steps and performing the necessary calculations, you can determine the height of the top of the trajectory, the impact speed, and the range of the projectile.