Two Dimensional Motion Problems - Physics

Summary

This video tutorial explores a two-dimensional motion problem involving a ball rolling off a 500-meter cliff at 5 m/s. It guides viewers through calculating the time to hit the ground, the horizontal range, and the final speed and velocity before impact using kinematic equations. The video also discusses projectile motion, free fall, and provides formulas for solving similar problems, encouraging viewers to check out additional resources for a deeper understanding.

Takeaways

• 📚 The video discusses a two-dimensional motion problem involving a ball rolling off a cliff.
• 🏞 The cliff is 500 meters high and the ball rolls off horizontally at 5 meters per second.
• 🕒 The time it takes for the ball to hit the ground is calculated using the kinematic equation for vertical motion.
• 📉 The initial vertical velocity (v y initial) is zero because the ball starts with only horizontal motion.
• 🌐 The vertical acceleration due to gravity is -9.8 m/s², which is used in the kinematic equation.
• 🔢 The calculated time for the ball to hit the ground is 10.1 seconds.
• 📏 The horizontal range or displacement of the projectile is found by multiplying the horizontal velocity (5 m/s) by the time of flight (10.1 s), resulting in 50.5 meters.
• 🚀 The final vertical velocity is calculated by adding the acceleration due to gravity over time, resulting in -98.98 m/s.
• 📐 The final speed of the ball before impact is found using the Pythagorean theorem, which yields 99.1 m/s.
• 🧭 To find the direction of the final velocity, the angle is calculated using the arctan function, giving an angle of 272.9 degrees from the positive x-axis.
• 📈 The video provides guidance on where to find additional resources, such as other videos on kinematics, projectile motion, and free fall physics.

Q & A

What type of motion is being discussed in the video?

The video discusses two-dimensional motion, specifically horizontal and vertical components of projectile motion.

What is the initial condition of the ball's motion in the video?

The ball rolls horizontally off a 500-meter cliff at a speed of 5 meters per second.

What is the formula used to calculate the time it takes for the ball to hit the ground?

The formula used is the kinematic equation for vertical motion: y_final = y_initial + v_y_initial * t + (1/2) * a * t^2.

What is the initial vertical velocity of the ball?

The initial vertical velocity of the ball is 0 meters per second, as it starts with only a horizontal component of velocity.

What is the acceleration acting on the ball in the vertical direction?

The acceleration acting on the ball in the vertical direction is the acceleration due to gravity, which is -9.8 m/s².

How long does it take for the ball to hit the ground?

It takes 10.1 seconds for the ball to hit the ground.

What is the range of the projectile in this problem?

The range, or horizontal distance the ball travels before hitting the ground, is 50.5 meters.

What is the final speed of the ball just before it hits the ground?

The final speed of the ball just before it hits the ground is 99.1 meters per second.

How do you calculate the final velocity of the ball?

The final velocity is calculated by combining the horizontal and vertical components of velocity using the Pythagorean theorem and considering the direction of the velocity vector.

What is the direction of the ball's final velocity just before it hits the ground?

The direction of the ball's final velocity is 272.9 degrees counterclockwise from the positive x-axis.

What additional resources are recommended for understanding two-dimensional motion problems?

The video recommends watching additional videos on kinematics, projectile motion, and free fall physics by Organic Chemistry Tutor on YouTube.

Outlines

00:00

📚 Introduction to Two-Dimensional Motion Problem

This paragraph introduces a two-dimensional motion problem involving a ball rolling off a 500-meter cliff at 5 meters per second. The goal is to determine how long it takes for the ball to hit the ground. The video suggests defining positions of interest and using kinematic equations to solve the problem. It also recommends watching additional videos on kinematics, projectile motion, and free fall physics for more context and formulas.

05:01

⏱ Calculating Time to Hit the Ground

The second paragraph focuses on calculating the time it takes for the ball to fall from the cliff to the ground. Using the kinematic equation for vertical motion, the problem is simplified by setting initial and final positions and identifying the relevant variables. The vertical acceleration due to gravity is considered, and the time (t) is calculated to be 10.1 seconds, which is the answer to the first part of the problem.

10:03

📏 Determining the Range of the Projectile

In this paragraph, the range of the projectile, which is the horizontal distance the ball travels before hitting the ground, is calculated. Since the horizontal velocity remains constant and the time of flight is known from the previous calculation, the range is found by multiplying the horizontal velocity (5 m/s) by the time (10.1 seconds), resulting in a range of 50.5 meters.

🚀 Finding the Final Speed Before Impact

The fourth paragraph addresses the calculation of the ball's final speed just before it hits the ground. The horizontal velocity remains constant, while the vertical velocity is determined by the acceleration due to gravity over time. The final speed is calculated using the Pythagorean theorem, combining the horizontal and vertical components of velocity, resulting in a final speed of 99.1 meters per second.

🧭 Calculating the Final Velocity and Direction

The final paragraph discusses determining the final velocity of the ball, which includes both magnitude (speed) and direction. The direction is found by calculating the angle using the arctan function of the vertical velocity over the horizontal velocity. The final velocity is described as 99.1 meters per second at an angle of 272.9 degrees counterclockwise from the positive x-axis, indicating the ball's trajectory in the fourth quadrant.

Keywords

💡Two-dimensional motion

Two-dimensional motion refers to the movement of an object in a plane, involving both horizontal and vertical components. In the video, the theme revolves around solving a physics problem where a ball rolls off a cliff, thus engaging in two-dimensional motion. The script discusses calculating the time it takes for the ball to hit the ground, illustrating the concept by defining positions of interest and using kinematic equations.

💡Projectile

A projectile is an object that is propelled into the air and moves under the influence of gravity alone. In the context of the video, the ball rolling off the cliff is considered a projectile, as it is only affected by gravity once it leaves the cliff. The script explains how to calculate the range or horizontal displacement of the projectile, which is a key aspect of understanding projectile motion.

💡Kinematics

Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause the motion. The video script references kinematics as it provides formulas and methods to solve for the time of flight, range, and final speed of the ball, which are all kinematic quantities. The term is used to guide viewers to additional resources for understanding motion problems.

💡Free fall

Free fall describes the motion of an object falling solely under the influence of gravity, starting from rest. The script mentions free fall when discussing the vertical acceleration of the ball as it falls from the cliff. It is a fundamental concept in the video's problem, as the ball's vertical motion is analyzed under the assumption of free fall conditions.

💡Velocity

Velocity is a vector quantity that represents the rate of change of an object's position with respect to time, including both speed and direction. The video script uses the term to describe the horizontal component of the ball's motion and later to calculate the final vertical velocity just before the ball hits the ground. It is essential in determining the ball's motion and final speed.

💡Acceleration

Acceleration is the rate of change of velocity of an object. In the video, the term is used to describe the constant acceleration due to gravity acting on the ball in the vertical direction, which is -9.8 m/s². This acceleration is crucial for calculating the time it takes for the ball to fall and its final vertical velocity.

💡Range

In the context of projectile motion, the range is the horizontal distance traveled by the projectile from its point of launch to the point of landing. The script explains how to calculate the range of the ball rolling off the cliff, which is a measure of the horizontal displacement during its flight.

💡Time of flight

Time of flight is the total time an object spends in motion from the point of launch to the point of landing. In the video, the script calculates the time of flight for the ball using the height of the cliff and the acceleration due to gravity, which is a critical step in solving the two-dimensional motion problem.

💡Speed

Speed is the scalar quantity that represents the magnitude of velocity, indicating how fast an object is moving without regard to its direction. The video script concludes with the calculation of the ball's final speed just before it hits the ground, using the Pythagorean theorem to combine the horizontal and vertical components of velocity.

💡Direction

Direction refers to the orientation or path along which an object is moving. In the video, the script discusses the final velocity of the ball, which includes both the magnitude (speed) and the direction of the ball's motion just before impact. The direction is determined by calculating the angle with respect to the horizontal using the arctan function.

💡Pythagorean theorem

The Pythagorean theorem is a principle in geometry that states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. The script applies this theorem to calculate the final speed of the ball by combining the horizontal and vertical components of its velocity.

Highlights

The video begins by introducing a two-dimensional motion problem involving a ball rolling off a cliff.

A ball is released horizontally from a 500-meter cliff at a speed of 5 meters per second.

The problem is broken down into parts to find the time it takes for the ball to hit the ground, the range, the final speed, and the final velocity direction.

The kinematic equation for vertical motion is introduced to calculate the time to hit the ground.

The initial vertical velocity (v_y initial) is zero due to the horizontal release.

The vertical acceleration is -9.8 m/s², representing the acceleration due to gravity.

The time calculation results in 10.1 seconds for the ball to reach the ground.

The range or horizontal displacement is calculated using the horizontal velocity and time.

The ball's range is determined to be 50.5 meters from the base of the cliff.

The final speed of the ball is calculated using the Pythagorean theorem, considering both horizontal and vertical velocities.

The final speed before impact is found to be 99.1 meters per second.

The final velocity direction is determined using the arctan function and the velocities' components.

The angle of the velocity vector is calculated to be 272.9 degrees from the positive x-axis.

The video provides additional resources for further study, including videos on kinematics, projectile motion, and free fall physics.

The video concludes by summarizing the steps taken to solve the two-dimensional motion problem.

The final velocity of the ball before hitting the ground is 99.1 m/s at an angle of 272.9 degrees.

Transcripts

in this video we're going to work on a

two-dimensional motion problem with

multiple parts

so let's go ahead and begin A ball rolls

horizontally off a 500 meter Cliff at a

speed of 5 meters per second

well let's begin with a picture

so let's say this is the cliff

and here is the ball

and it's going to roll off the cliff and

then it's going to hit the ground

how long will it take for the ball to

hit the ground

now what I like to do is I like to

Define

the positions of Interest we'll call

this position a which is the initial

position and position B

or the final position

now we know the height of the cliff it's

500 meters

and we know

the ball has a horizontal speed

of 5 meters per second

let's write that better

so with this information how can we find

the time it's going to take

from for the ball to go from position a

to position B

how long will it take for it to hit the

ground

now hopefully you have

a list of physics equations with you if

not

I recommend that you go to YouTube and

in the search bar type in kinematics

organic chemistry tutor I have a video

entitled kinematics which will give you

a list of formulas that you need to

solve two-dimensional motion problems

even one dimensional motion problems as

well

also check out my video projectile

motion

that's going to be more of the

two-dimensional motion problems and it's

going to have even more formulas for you

for you to use when solving uh

these types of problems

but the equation that we could use is

this one since we're dealing with

vertical motion

y final is equal to Y initial Plus

v y initial

t

plus one half

a t squared

now there's a lot of variables in this

formula our goal is to calculate t

now what is y initial and Y final

if we assume this to be ground level

that means position B

is at a y value of zero so that's going

to be y final

B is the final position a is the initial

position y initial

well that's going to be the height of a

cliff that's at 500 meters

so we have that

now what about v y initial

now the ball initially is moving

horizontally it only has an X component

it doesn't have a y component

so for any object moving in horizontal

in a horizontal Direction

the vertical velocity is going to be

zero

so that's v y initial at Point a

now the vertical acceleration

for any object in free fall it's always

going to be negative 9.8

because that object is under the

influence of gravity so any object

on Earth under free fall it's going to

have that

vertical acceleration

by the way you should check out my other

video If you go to YouTube and type in

free fall physics organic chemistry

tutor I have another video that talks

about

how to solve problems

with objects in free fall

including objects like this one

so feel free to check out those three

videos free fall physics kinematics and

projectile motion for those of you who

want more problems

so now let's

finish this problem why filo is zero y

initial is 500

v y initial is zero

and then it's plus one half the

acceleration is negative 9.8 and we need

to solve for t

so moving to 500 to the other side

we're going to get Negative 500

one-half times negative 9.8 that's going

to be negative 4.9

so now what we need to do is divide both

sides by

negative 4.9

these will cancel

and we'll get t squared is equal to

negative 500 divided by negative 4.9

that's going to be

102.04

taking the square root of that number

this will give us 10.1

so let's put that here it's going to

take

10.1 seconds

for the ball to go from position a to

position B

so that's the answer for the first part

of the problem

now let's move on to Part B

how far from the base of the cliff will

the ball land

in other words what is the range of this

projectile

a projectile is simply an object

under the influence of gravity

in fact gravity is the only force acting

on a projectile

so a rocket would be considered a

projectile because

it's affected by the thrust of the

engine but any object that's moving in

air only under the influence of gravity

by definition

is a projectile

so how can we calculate the range

of this projectile

the range is simply the horizontal

distance

or you could say horizontal displacement

in this problem they're the same

since the object doesn't change

direction

so the range is going to be DX

and if you want to find the displacement

it's equal to the velocity multiplied by

the time so if we want to find a

horizontal displacement it's a v x times

t

now for a projectile the horizontal

velocity is constant the vertical

velocity changes with time because the

object has a vertical acceleration

but if you want to just find the range

it's simply equal to VX times T in this

example VX is 5 meters per second

and we know the object is going to be

moving for 10.1 seconds

based on the answer that we got in part

a

so 5 times 10.1

that gives us a range of 50

.5 meters

that's the answer for Part B

now what about part C

what is the final speed of the ball just

before it hits the ground

now at this point we need to realize

that

the ball doesn't just have an X

component

its velocity has a y component as well

the X component is going to remain

constant it's not going to change

because the acceleration in the X

direction is zero

however the vertical velocity of the Y

component will change because there is

an acceleration in the y direction

to find the New Vertical Velocity at

point B

it's going to be equal to the initial

Vertical Velocity

plus a t so this is the formula V final

is equal to V initial plus a t

just in the y direction

v y initial is zero

the acceleration in the y direction is

negative 9.8

and the time that this object is going

to be in Flight is 10.1 seconds

negative 9.8 times 10.1

and that gives us a vertical velocity

of negative 98.98

meters per second

so right now

the ball at point B just before it hits

the ground

has a horizontal velocity of 5 meters

per second

and it has a vertical velocity

of negative

98.98 meters per second

so this is VX and this is v y

our goal is to find the final speed so

we need to get V

using the Pythagorean theorem

V is going to be the square root of VX

squared plus v y squared

a squared plus b squared is equal to c

squared if you want to calculate C the

hypotenuse is equal to the square root

of a squared plus b squared

so v x is five

and v y is negative 98.98

so go ahead and plug that in

so V is 99.1

meters per second

this is the answer to part C and that is

the final speed speed

is always positive velocity can be

positive or negative

now Part D is asking for the final

velocity of the ball just before it hits

the ground whereas part C is asking for

the final speed

we have the final speed

but to get the final velocity

we need the direction remember velocity

is speed with Direction

we have the magnitude which is the speed

but we need to get the direction in

other words we need to get the angle

so let's calculate the angle Theta

the angle inside of that acute triangle

is going to be the reference angle

and to find it we're going to take

arctan

of v y over VX

now we're only going to use the absolute

values of v y and VX we don't need to

worry about the negative sign right now

so it's 98.98 over 5.

this gives us a reference angle

of 87.1 degrees

but I'm going to get the angle measured

counterclockwise from the positive

x-axis

that's going to be 360 minus the

reference angle

so 360 minus 87.1 will give us an angle

of

272.9 degrees which tells us that

the vector the velocity Vector is in

quadrant four it's moving in that

direction

so part C the speed the final speed just

before hits the ground is simply 99.1

meters per second part D the final

velocity is the combination of these two

answers

the final velocity of the ball just

before histogram is 99.1 meters per

second at an angle of 272.9 degrees

counterclockwise

from the positive x-axis

so that's basically it for this video

so now you know how to solve a typical

two-dimensional motion problem

so if you want more problems like this

check out my videos on projectile motion

kinematics and Free Fall physics