Motion in Two Dimensions - AP Physics 1

First, find the horizontal (x) and vertical (y) components of the velocity

Next, find how long the object is in the air by calculating the time it takes it to reach the top of its path, and doubling that number.

t = 6.25s

Total time in the air is therefore 12.5s (twice this value).

Finally, find distance traveled my multiplying horizontal velocity and time.

We can start this problem by determining how far the box will travel on the ramp before hitting the ground. Since the vertical distance is 30m and we know the angle of the ramp, we can determine the length of the hypotenuse using the equation .

Now that we have the distance traveled, we can determine the acceleration on the box due to gravity. Because the box is on a sloped surface, the box will not experience the full acceleration of gravity, but will instead be accelerated at a value of . Since the angle is 60o, the acceleration on the box is

Finally, we can plug these values into the following distance equation and solve for time.

We can start this problem by determining how much time it takes the box the reach the ground. Since the vertical distance is 30m and we know the angle of the ramp.

Now that we have the distance traveled, we can determine the acceleration on the box due to gravity, using the equation .

We can plug these values into the following distance equation and solve for time.

Now that we know the acceleration on the box and the time of travel, we can use the equation to solve for the velocity.

This is a two step problem. The first step is to calculate the time it takes for the ball to reach the ground. To find this time, we use the following kinematic equation dealing with vertical motion.

Choosing the ground to be the zero height, we have and .

Also, knowing that the initial vertical velocity is zero, we know that .

The kinematic equation simplifies using these values.

Rearrange the equation to isolate time.

We know that is the acceleration due to gravity: . Plug in the values to solve for time.

We now have the time the ball is travelling before it hits the ground. Use this value to find the horizontal distance before it hits the ground with the kinematic equation .

We know that and that . Using these values and the time, we can solve for the horizontal distance travelled.

First, determine how far the car travels in each direction:

Now that we have the directional displacements, we can find the total displacement by using the Pythagorean Theorem.

Find the average velocity by dividing the total displacement by the total time.

Velocity is a vector, meaning it has both magnitude and direction. Now that we have the magnitude, we can find the direction by using trigonometry.

Use the north and the east directional displacements to find the angle.

Our final answer will be:

Any projectile has a vertical velocity of zero at the peak of its flight. To solve this question, we need to find the time that it takes the ball to reach this height. The easiest way is to solve for the initial vertical velocity using trigonometry, and then use the appropriate kinematics equation to determine the time.

We know that the final vertical velocity will be zero. We can solve for the initial vertical velocity using the given angle and total velocity.

Using this in our kinematics formula, we solve for the time.

Keep in mind that this is only the vertical velocity. The total velocity at the peak is not zero, since the ball will still have horizontal velocity.

First we need to find out at what the speed relative to the ground the plane needs to fly. The plane needs to cover 100 miles in 1 hour, so it's simply 100 mph due north.

Now we need to calculate its speed relative to the wind and its angle. We know the sum of the wind vector and plane vector needs to equal 100 mph due north. Therefore, all east/west movement must cancel out and all north/south movement must add to 100mph.

We can separate the wind velocity into its components:

We can also represent the components of the plane's velocity:

If the trig functions seem reversed, this is because the angle in question is between the y-coordinate and a vector pointing slightly west of north.

We can also represent the components of the velocity relative to the ground:

Since the horizontal velocity is equal to 0, we can set the x-components of the wind and plane vectors equal to each other:

The sum of the y-components of the wind and plane vectors must equal 100:

The wind vector is subtracted because it is in the opposite direction of the plane vector.

Now we just need to isolate a variable and substitute one equation into another. We will isolate the total plane velocity in the first equation:

Substituting this into the second equation, we get:

Solving for theta, we get (west of north)

We can plug this into the first equation to get:

To solve this problem, we first need to split the velocity into its vector components.

Initial vertical velocity:

Initial horizontal velocity:

Since we are neglecting air resistance, horizontal velocity does not change over time. We only need to calculate the new vertical velocity after two seconds, using kinematics:

The negative sign simply means that the vertical velocity has changed direction, and is now pointed downward.

We can use the following equation to determine the total velocity, which will be the sum of the horizontal and vertical velocity vectors:

If , the horizontally thrown ball will travel no distance. As the height of the cliff increases, the horizontally thrown ball will have more and more air time until it begins to travel further than the ball thrown above the horizontal. This is because the distance each ball will go equals the time it's in the air multiplied by its horizontal velocity (since it's not accelerating in the x-direction). The time the ball thrown above the horizontal is always the same amount greater than the time the ball thrown horizontally is in the air. The ball thrown horizontally has a horizontal velocity greater than the ball thrown above the horizontal (by a factor of ). This means the ball thrown horizontally travels a distance of while the ball thrown above the horizontal travels a distance of . If we equate these two expressions, we get . When is less than this, the ball thrown above the horizontal travels further and when is greater the ball thrown horizontally travels further. represents the time is takes for the ball thrown horizontally to hit the ground when the height of this cliff is .

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.

How much time it takes the ball to reach its highest point depends on its motion in the y-axis. Therefore, it depends on the ball's initial velocity in the y-direction and the acceleration of gravity. Note that at the highest point of its trajectory, the ball reaches a velocity of in the y-direction for an instant, just before it changes direction and starts coming down. Therefore we can use the following kinematic equation for motion with constant acceleration to find :

Here, is the final velocity on y for this part of the ball's trajectory (i.e. its velocity at the highest point), is the ball's initial velocity in the y-direction, and is the acceleration acting on the y-axis (gravity). We can express in terms of and by obtaining the y component of the initial velocity of the ball, which is given by:

Note that the acceleration in y-direction is negative since gravity pulls downwards.

Solve for .

An important thing to notice is that this projectile motion is symmetrical, meaning that the ball starts and ends its motion at the same height. Then, it must be the case that the ball takes the same amount to reach its highest point from its original position as it takes the ball to fall down from the highest point to its final position. In other words, it takes a time of to complete the entire trajectory. Since the ball's movement follows projectile motion, it experiences motion with constant velocity on the horizontal axis (the x-axis). We can calculate this velocity by obtaining the x-component of the ball's initial velocity:

Since the velocity is constant, we can express it in terms of displacement in the x-direction and time with the following equation:

Here, is the displacement in the x-direction and is the time.

Therefore, we have that for the range (the ball's entire displacement in the x-direction for its motion) is given by:

We're told that the golf ball is starting its parabolic journey with a certain velocity at an angle to the ground. To solve for the time the golf ball stays airborne, we'll need to consider the x and y-components of the ball's trajectory.

First, we'll need an expression that considers the ball's velocity in the x-direction.

We'll also need an expression for the y-component of the velocity.

We'll need an expression that can relate the vertical distance traveled with the time spent in the air. Since the only acceleration occurring in this scenario is due to gravity, the result is that acceleration is constant. Therefore, we can make use of some of the kinematics equations:

It's also important to note that once the ball lands back on the ground, only its horizontal displacement will have changed, while its vertical displacement will remain unchanged.

We also have to remember that in this case, the source of acceleration is from gravity, which points downwards. If we define up as the positive y direction, then down must be the negative y direction. Therefore, we can write:

Plug in the vertical component of velocity and solve for time.

In order for the object not to gain velocity, all the forces acting on it need to cancel out. The normal force will already cancel with the perpendicular force due to gravity () by definition. In order to find the correct coefficient of friction, we set the magnitude of the force due to friction equal to that of the parallel force due to gravity :

The mass and gravity cancel here.

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.

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.