This problem is greatly simplified since the home run fence has the same height as the initial height of the ball. Therefore, we can use this range equation to solve this problem in one calculation: 

Rearranging for velocity, we get:

Plugging in our values, we get:

There are multiple ways to attack this problem, so don't worry if you took a different route. However, no matter which way you go, you'll be substituting small expressions into larger ones.

First, let's develop an expression of how long it takes the ball to reach the home run fence. We know that:

Rearranging for time:

Then we can substitute the following expression in for horizontal velocity:

Now that we have an expression for time, we can use the following expression:

Plugging in an expression for initial vertical velocity, we get:

Plugging in our expression for time, we get:

Now let's simplify this a bit:

Rearranging:

And some more:

One last time:

It may look a bit nasty, but we know all of these values, so time to plug and chug:

From the problem statement, we can use the following piecewise function to express the effective "height" at any point within the stadium:

If      then   

If      then   

Before, moving on, make sure you understand the reasoning of this function. The wall begins at a distance of , a height of , and rises at an angle of .

Using the total horizontal distance traveled, we can calculate the ball's final height using the above expression:

Now we can use the following two kinematics expression:

Where d is the horizontal distance traveled. Replacing out component velocities, we get:

Substituting our expression for time into the other expression, we get:

Simplifying, we get:

Rearranging:

And again:

One last time:

It's nice and messy, but we have all of these values, so time to plug and chug:

From the problem statement, we can use the following piecewise function to express the effective "height" at any point within the stadium:

If      then   

If      then   

Before, moving on, make sure you understand the reasoning of this function. The wall begins at a distance of , a height of , and rises at an angle of .

We can use the following expression to calculate the final height of the ball:

Plugging in our values, we get:

Then we can use the piecewise expression to determine to total horizontal distance traveled:

From the problem statement, we can use the following piecewise function to express the effective "height" at any point within the stadium:

If      then   

If      then   

Before, moving on, make sure you understand the reasoning of this function. The wall begins at a distance of , a height of , and rises at an angle of .

We can quickly determine the horizontal distance traveled by the ball:

Then we can use the piecewise function to calculate the final height of the ball. Since , we will use the second expression:

Then we can use the following kinematics equation:

Rearranging, we get:

Plugging in our values, we get:

Then we can use the following expression to determine the initial angle of the ball:

Rearranging and plugging in our values, we get:

From the problem statement, we can use the following piecewise function to express the effective "height" at any point within the stadium:

If      then   

If      then   

Before, moving on, make sure you understand the reasoning of this function. The wall begins at a distance of , a height of , and rises at an angle of .

We can calculate the total horizontal distance traveled by the ball using the following expression:

The ball has yet to reach the home run wall, so it will land back on the ground with a final height of .

We know that the brick is dropped from rest  at the top of the building and, ignoring air friction, we know the only relevant force is due to gravity .  Using the kinematic equation , plug in our known values for , , and , then solve to get .

The problem can be split into two parts: when the brick is in the air above the building, and when the brick begins its descent past the top of the building once more. Since we are finding the time the brick takes to travel in the vertical direction, only the y-component of the velocity is relevant, .

To find the time the brick is in the air for the first part, use the kinematic equation . The time the brick takes to reach its peak is given by  , so . Doubling this  gives the time the brick takes to rise then fall again to the height from which it was thrown, so .

To find the time the brick takes to fall from the top of the building to the ground, use the kinematic equation . We know that the initial velocity for this part, which is now pointing downward, is equal in magnitude to when the brick was first thrown, . The height of the building is given by . Hence our equation becomes . Solving for  gives .

Adding the two times together, .

In projectile motion, there are no horizontal forces acting on the object during its flight. By Newton's Second Law we know that if no force is applied to an object, it does not accelerate.

So, the cannonball's horizontal velocity remains constant. Twice the time of flight leads to twice the horizontal distance.

Determining the time for the upward velocity to equal zero:

The ball will take the same amount of time to descend back to it's initial height, the height of Jennifer.

Determining how far horizontally the ball will travel in that time.