Start by drawing a picture.

By traveling both north and east, Peter has displacement along both the x-axis and the y-axis. Recognize that this makes a right triangle.

We can solve for the x-axis displacement by using trigonometry.

In this case, we know the angle and the hypotenuse. The x-displacement is adjacent to the angle.

Multiply both sides by  and solve.

Since we are solving for the displacement, we need to include the direction. The displacement is .

When a projectile is launched at an angle, the result is an initial velocity with both horizontal and vertical components.

Once these initial components are determined, we can find how they change during flight. The only thing capable of changing a given velocity is an outside force (Newton's first law). The only force acting on the ball in the air is gravity, which serves to decrease the positive vertical velocity. There are no forces in the horizontal direction. Since gravity acts perpendicular to the horizontal velocity, it has no effect and cannot change the horizontal velocity value. As a result, the horizontal velocity remains constant at all points during projectile motion, even though the vertical velocity will change due to gravity.

This problem involves understanding motion in two dimensions: horizontal (we will use ) and vertical (we will use ).

Let's begin by writing down what we know, and what we need to find.

What we know:

What we need to find:

To find velocity, we need to know both the horizontal and the vertical components of the ball's velocity. We will break them down below.

The horizontal component of the velocity, , does not change with time so we can solve for that simply by finding the horizontal distance, , covered over time .

The vertical component of the velocity, , will be slightly more challenging to find. Because gravity acts in the vertical direction, we now need to take acceleration into account. Because of this, we will try to find the initial vertical velocity, , of the ball. We are told in the problem that the ball hits the scoreboard at the top of its trajectory. At the top of the trajectory, vertical velocity will be zero. This also marks the midpoint of the ball's flight, or half the given time. With this information, we can solve for the initial vertical velocity:

Now that we have two components, we can use simple trigonometry to find the angle. In a right triangle made by the height of the scoreboard, the ball's trajectory, and the ground, the tangent of the angle will be equal to the vertical component divided by the horizontal component. This means that we can find the tangent of the angle by dividing vertical velocity by horizontal velocity:

We will use horizontal and vertical kinematics to solve this problem. We will begin by writing down what we know:

We want to know:

To begin, let's discuss how we will find the take-off speed. It would be simple to calculate  if we had one of the vector components, say .

              (Equation 1)

While we are given the angle , we do not have . So to find that, we use an equation:

                 (Equation 2)

Now we know , but we don't have time , so we will try to find that using our vertical component. We know that the time it takes for the jumper to reach his peak height will be half the total time. We also know that at the peak height vertical velocity will be zero.

                (Equation 3)

Now, we don't know what  is, but we do know its relation to  is:

Now, let's plug that into equation 3:

We can now plug this into equation 2:

Solve this equation to isolate .

We can now plug in our given values for horizontal distance, angle, and acceleration to solve for the horizontal velocity.

Now we can plug that back into equation 1:

This problem involves using vectors and trigonometry. We are told that the swimmer is first seen  away at a  angle to the fan's right. If we were to draw that below, that would look like this:

The black line represents the fan's line of sight directly in front of him while the red line represents the line of sight from the fan to his favorite swimmer. We are then told that  later the swimmer is seen away at a  angle to the fan's left. We will add that line of sight to our drawing in green:

Now we can use trigonometry to find how far the swimmer needed to swim in those . We can break up the swim into two parts. The first part, , is when the swimmer is to the right of the fan until he is directly in front of the fan. That is represented by the thin horizontal red line below. The second part, , starts when the swimmer is directly in front of the fan until he swims to the left of the fan, represented by the thin horizontal green line.

We can find each of these distances using trigonometry, since we know the angles in the right triangles. To find the distance swam in the first part, we use sine.

We can also use sine to find the distance in the second part:

To find the total distance swam, we add this distance from each part.

We also know that it takes the swimmer  to cover this distance. To find how fast he is swimming, we use the definition of velocity:

This problem involves understanding motion in two dimensions: horizontal (we will use ) and vertical (we will use ).

Let's begin by writing down what we know, and what we need to find.

What we know:

What we need to find:

To find velocity, we need to know both the horizontal and the vertical components of the ball's velocity. We will break them down below.

The horizontal component of the velocity, , does not change with time so we can solve for that simply by finding the horizontal distance, , covered over time .

The vertical component of the velocity, , will be slightly more challenging to find. Because gravity acts in the vertical direction, we now need to take acceleration into account. Because of this, we will try to find the initial vertical velocity, , of the ball. We are told in the problem that the ball hits the scoreboard at the top of its trajectory. At the top of the trajectory, vertical velocity will be zero. This also marks the midpoint of the ball's flight, or half the given time. With this information, we can solve for the initial vertical velocity:

Now that we have both the horizontal and vertical components, we can use the Pythagorean theorem to solve for the total initial velocity, .

The key to solving this problem is to recognize that in two dimensional motion, the x-direction and the y-direction are dealt with separately. The question asks how long the projectile is in the air so we are dealing only with motion in the y-axis. 

Since the projectile was fired at an angle we must use trigonometry to find the velocity in the y-axis . Then we use a kinematic equation to solve for time.

Initial velocity in the y-axis:

Now we will find the time it takes for the projectile to hit the ground using a kinematic equation:

We use at the top of the projectile's trajectory, where its velocity is changing from going up, to going down. 

Note that this is the time it takes for the projectile to reach maximum height. The time it takes to get back to the ground from here is the same as the time it took to get here. Thus, to find the total flight time, we must double our answer.

The question asks how far the projectile can travel before landing. Since we deal with the x and y-axes of 2-dimensional motion separately, it is important to recognize that only the movement in the x-axis is relevant. If you are driving 50mph and do so for three hours you have driven 150mi. Similarly, our projectile is traveling at a certain speed horizontally and does so for 20s. Thus we simply multiply its horizontal speed by the time the projectile was in the air. Since the projectile was launched at an angle we use trigonometry to find its horizontal speed:

Now use the formula for range and plug in the velocity in the x-direction (constant) to find the distance travelled.

Acceleration is defined by the rate of change of velocity per unit time. Thus, if the acceleration of gravity on Earth is  (which is an approximation close enough to the actual value for our purposes) then all objects are accelerated downward by 10 meters per second every second. Since the ball was thrown directly upward and will fall straight down, we need not worry about sideways motion and can focus on gravity slowing the ball until at reaches a maximum height and falls back to Earth. The ball was thrown upward with an initial velocity of . Gravity will slow the ball down by  until it reaches a maximum height. Thus, the ball will reach this maximum height in 2.2s. Mathematically: 

It then falls back to Earth for 2.2s. Thus, the ball takes a total of 4.4s to reach the ground.

The question gives the total initial velocity, but we will need to find the horizontal component using trigonometry.

To find the horizontal velocity we use the equation .

We can plug in the given values for the angle and initial velocity to solve.