The cannon ball has a two-dimensional motion, specifically projectile motion. Any two-dimensional motion consists of two one-dimensional motions happening at the same time. This is why we break vectors into components, so we can analyze each one-dimensional motion separately. The motion in the y-axis and the x-axis are connected by the time variable, since they happen at the same time. For projectile motion, we have that on the y-axis (the vertical axis) a constant acceleration (gravity) is acting on the cannon ball. For the y-axis motion with constant acceleration or free fall (which is motion with constant acceleration where the acceleration is gravity) are correct choices. On the x-axis (the horizontal axis), there is no acceleration acting on the cannon ball horizontally. Therefore, on the x-axis the cannon ball experiences motion with constant velocity. Then the only answer choice that describes the motion of the cannon ball in each axis correctly is: y-axis: motion with constant acceleration and x-axis: motion with constant velocity.

At the projectile's maximum height, we know that it's vertical velocity will be  Since the only type of acceleration the object faces is the acceleration due to gravity, we can model its velocity using only two terms: it's initial velocity and acceleration due to gravity.  Recall that the formula for velocity with constant acceleration is given as:

Here,  is the object's constant acceleration,  is the initial velocity, and  is time. 

The only acceleration is acceleration due to gravity, which is:

To determine initial velocity , we have to determine a few other things. The object is being launched at a speed of  at an angle of . Therefore, the vertical component (the vertical velocity of this object) is going to be 

Therefore, we can model the velocity of this object as:

The signs of the two terms must be opposite because they are going in different directions. Negative velocities in this model mean the object is moving towards the center of the earth. 

Since the object reaches its maximum point when  we must solve the following equation for 

Superman needs to cover 3,000 meters. But the question is how much time does he have to cover those 3,000 meters? If you know the time, you can just divide 3,000 meters by the time he is traveling and you have his speed! But how do we know how quickly he needs to cover the 3,000 meters? The falling cat tells us! The acceleration due to gravity is a constant on the surface of Earth and it will dictate how quickly something hits the ground once dropped. Considering we ignore air resistance in physics 1, we don't care about the object's size or mass. So all we need to do is find the appropriate kinematic equation(these equations relate displacement, velocity, acceleration and time) and solve for time. The way I chose a kinematic equation is I organize variables into things we are told in the problem and things we are not told in the problem. In this case, we are given information about the height of the cat, it's mass and that it slipped (velocity in both x and y directions is zero). Not much to go off of but it is all we need! We are solving for time and we have information about initial velocity and displacement.

The equation: 

We can see with this equation that under the assumption that the initial velocity is zero, it simplifies to a relationship between x and t (acceleration is g and is a constant). Rearranging for t gives: 

Plugging in the appropriate numbers, the time that it take for the cat to hit the ground will be 1.43 seconds. Now we know Superman must cover the 3,000 in 1.43 seconds.

Dividing 3,000 by 1.43 yelids a velocity of  

. 

At first this problem could seem challenging, but in fact, it is straight-forward if you can remove the initial velocity from the equations.  We begin by writing the equations of motion for the x and y directions.

We then devide the two equations to get an expression for the tangent of the angle, as well ass remove the initial velocity from the problem.  This enables us to use the information given to calculate the angle the projectile was fired.

The top of the hill is  above the ground. However, that isn't the distance from the bottom of the hill to the top, since the car is going up an incline of . 

The length of the incline can be given as: 

, where  is the length of the incline. 

Solving for 

. 

The incline is  in length. 

At a speed of , 

First we need to find the vertical velocity component (y-component). We know our velocity is  at an angle of . We need to use the sine function to determine the y-component:

Now we use the kinematic equation to determine the height reached:

We solve for  which is our height and substitute our known values.

Our initial vertical velocity is  and final vertical velocity is  at the peak.

Since the question is asking how long the projectile was airborne, we are dealing with motion in y-direction only. Find initial velocity in the y-direction and substitute into the kinematic equation:

Narrow the choices of kinematic equations by including only ones with distance in their formulas. Of those two, we want the one that includes time:

The red term becomes zero because our projectile is not moving at first. Plug in known values and solve.

Find the time the object is airborne by solving for t in the equation:

Then, multiply the time in the air by the initial velocity to get the range of the projectile. Note that the initial velocity is zero in the y-direction.

This is a two-part problem. First, we need to know how fast the block is moving in the horizontal direction when it leaves the table. This can be calculated by using a kinematic equation, combined with Newton's second law.

Now that we know how fast the block is moving horizontally, we need to know how long the block will be in air before hitting the ground. This is found by considering the equations of motion for an object falling from rest:

Lastly, we find the horizontal distance by: