All High School Physics Resources
Example Questions
Example Question #47 : Linear Motion
A cannon on level ground fires a cannon ball at at above the horizontal. How long does it take the ball to reach its maximum height?
To find the time for the projectile to reach its maximum height, we can use the appropriate kinematics equation:
We know that the final velocity at the maximum height will be zero, and we also know the acceleration due to gravity. Before we can use the equation, however, we must solve for the initial vertical velocity. We are given the total initial velocity and the angle of the initial trajectory. Using these values, we can use trigonometry to solve for the initial vertical velocity.
Now that we know the initial vertical velocity, we can return to the kinematics equation to solve for the time.
Example Question #11 : Understanding Motion In Two Dimensions
A cannon on level ground fires a cannon ball at at above the horizontal. What is the total time the ball is in the air?
To find the total time, we can solve for the time to reach the peak and then double this value to find the total time.
To find the time for the projectile to reach its maximum height, we can use the appropriate kinematics equation:
We know that the final velocity at the maximum height will be zero, and we also know the acceleration due to gravity. Before we can use the equation, however, we must solve for the initial vertical velocity. We are given the total initial velocity and the angle of the initial trajectory. Using these values, we can use trigonometry to solve for the initial vertical velocity.
Now that we know the initial vertical velocity, we can return to the kinematics equation to solve for the time.
Double this value to find the total flight time.
Example Question #12 : Understanding Motion In Two Dimensions
A cannon on level ground fires a cannon ball at at above the horizontal. What is the total horizontal distance that the cannon ball travels?
To solve for the horizontal distance we can use the simple formula for velocity, since there is no acceleration in the horizontal direction.
This leaves us two distinct variables: and .
To find our , we need to use cosine.
To find the time, we can solve for the time to reach the peak and then double this value to find the total time.
To find the time for the projectile to reach its maximum height, we can use the appropriate kinematics equation:
We know that the final velocity at the maximum height will be zero, and we also know the acceleration due to gravity. Before we can use the equation, however, we must solve for the initial vertical velocity. We are given the total initial velocity and the angle of the initial trajectory. Using these values, we can use trigonometry to solve for the initial vertical velocity.
Now that we know the initial vertical velocity, we can return to the kinematics equation to solve for the time.
Double this value to find the total flight time.
Now that we know our total time and horizontal velocity, we can return to the velocity equation to solve for the distance traveled.
Example Question #43 : Motion And Mechanics
Peter walks north of east. What is his total displacement along the y-axis?
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 y-axis displacement by using trigonometry.
In this case, we know the angle and the hypotenuse. The y-displacement is opposite the angle.
Multiply both sides by and solve.
Since we are solving for the displacement, we need to include the direction. The displacement is .
Example Question #51 : Linear Motion
Peter walks north of east. What is his total displacement along the x-axis?
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 .
Example Question #13 : Understanding Motion In Two Dimensions
A ball is shot out of a cannon angled above the ground. At what point is the ball's horizontal speed the greatest?
Half way between release from the cannon and the maximum height of trajectory
The instant before the ball hits the ground
Immediately after release from the cannon
It remains constant
At the maximum height of trajectory
It remains constant
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.
Example Question #51 : Motion And Mechanics
An athlete kicks a ball into the air. It travels and is in the air for . One stadium has a large scoreboard in the center of the field that reaches above the ground. With what angle must the ball leave the athlete's foot in order to just barely clear the height of the scoreboard?
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:
Example Question #52 : Linear Motion
A long-jumper jumps at a angle and lands away. What is the take-off speed of the jumper?
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:
Example Question #11 : Understanding Motion In Two Dimensions
At a swim meet, a fan is eagerly awaiting his favorite swimmer. As the race begins the fan spots his favorite swimmer away at a angle to his right. later the swimmer is away at a angle to his left. How fast is the swimmer swimming?
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:
Example Question #18 : Understanding Motion In Two Dimensions
An athlete kicks a ball into the air. It travels and is in the air for . One stadium has a large scoreboard in the center of the field that reaches above the ground. With what minimum total velocity must the athlete kick the ball in order to get it over the scoreboard?
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, .