Two Dimensional Motion (4 of 4) Horizontal Projection, Worked Example

Summary

In this educational video, the host explains a two-dimensional projectile motion problem. An object is projected horizontally from a height of 45 meters with an initial velocity of 37 m/s. The video focuses on calculating the horizontal distance traveled by the object. The host clarifies that motion in the X and Y directions is independent, with zero acceleration in X and gravitational acceleration in Y. Using kinematic equations, the time of flight is determined to be 3.03 seconds, leading to a horizontal distance of 112 meters. The explanation is designed to help viewers understand the principles of projectile motion.

Takeaways

• 🎯 The problem involves calculating the horizontal distance traveled by an object in two-dimensional projectile motion.
• 📏 The object is projected from an initial height of 45 meters with an initial horizontal velocity of 37 m/s.
• 🔄 The motion in the X and Y directions are independent of each other.
• 🚫 In the X direction, there are no forces acting on the object, resulting in zero acceleration.
• 🌐 In the Y direction, the only force acting is gravity, causing a free-fall motion with an acceleration of -9.81 m/s².
• 📐 The kinematic equations are used to solve the problem, involving variables such as initial and final velocity, change in position, acceleration, and time.
• ⏱ The time taken for the object to move in the X direction is the same as the time taken to fall in the Y direction.
• 🔢 The time for the object to fall in the Y direction is calculated using the kinematic equation and the given values.
• 📉 The change in position in the Y direction is -45 meters, indicating a downward movement.
• 📏 The horizontal distance traveled in the X direction is calculated by multiplying the horizontal velocity by the time taken.

Q & A

What is the initial height from which the object is projected?

The object is projected from an initial height of 45 meters.

What is the initial velocity of the object in the horizontal direction?

The initial velocity of the object in the horizontal direction (X direction) is 37 meters per second.

What is the acceleration in the X direction?

The acceleration in the X direction is zero meters per second squared because there are no forces acting on the object in that direction.

What is the acceleration in the Y direction?

The acceleration in the Y direction is equal to the acceleration due to gravity on Earth, which is -9.81 meters per second squared.

What is the initial velocity in the Y direction for free fall motion?

The initial velocity in the Y direction for free fall motion is 0 meters per second.

How do we determine the time it takes for the object to fall in the Y direction?

We use the kinematic equation that relates change in position, initial velocity, time, and acceleration: \( \Delta y = v_{iy} \cdot t + \frac{1}{2} a_y \cdot t^2 \), where \( v_{iy} \) is the initial velocity in the Y direction, \( a_y \) is the acceleration in the Y direction, and \( \Delta y \) is the change in position in the Y direction.

What is the change in position in the Y direction?

The change in position in the Y direction is -45 meters, indicating the object moves downward from its initial height.

How is the time it takes to fall in the Y direction related to the time in the X direction?

The time it takes for the object to fall in the Y direction is the same as the time it takes to move horizontally in the X direction because the motion is simultaneous and independent in both directions.

How far does the object travel in the X direction?

The object travels 112 meters in the X direction.

How long does it take for the object to travel the distance in the X direction?

It takes 3.03 seconds for the object to travel the distance in the X direction.

What are the key steps to solve this projectile motion problem?

The key steps include identifying the acceleration in both the X and Y directions, using the kinematic equations to find the time of flight in the Y direction, and then using that time to calculate the distance traveled in the X direction.

Outlines

00:00

📐 Introduction to 2D Projectile Motion

The video begins with an introduction to a physics problem involving two-dimensional projectile motion. The goal is to calculate the horizontal distance an object travels when projected from a height of 45 meters with an initial horizontal velocity of 37 m/s. The presenter emphasizes that the motion in the horizontal (X) direction and the vertical (Y) direction are independent of each other. In the X direction, there is no acceleration due to balanced forces, while in the Y direction, the object experiences free fall with an acceleration of -9.81 m/s² due to gravity. The presenter outlines the need to use kinematic equations to solve the problem and identifies the known variables: acceleration in both directions, initial velocities, and the change in position for the Y direction. The unknowns to be determined are the final velocities and the change in position in the X direction. The time of flight is also unknown but is crucial for solving the problem.

05:02

🕒 Solving for Time and Horizontal Distance

In the second paragraph, the presenter focuses on solving for the time it takes for the object to fall in the Y direction, which is also the time it travels in the X direction due to同步运动. Using the kinematic equation that relates change in position to initial velocity, acceleration, and time, the presenter simplifies the equation by acknowledging that the initial velocity in the Y direction is zero. This simplification leads to a direct calculation of time using the formula: time = √(2 * change in position in Y / acceleration in Y). Plugging in the values, the time is found to be 3.03 seconds. With the time known, the presenter then calculates the horizontal distance traveled in the X direction by multiplying the horizontal velocity (37 m/s) by the time (3.03 seconds), resulting in a distance of 112 meters. The presenter concludes by summarizing the key points: the acceleration in the Y direction is due to gravity, the acceleration in the X direction is zero, and the time of travel in both directions is the same. The final answer is that the object travels 112 meters in the X direction in 3.03 seconds.

Keywords

💡Two-dimensional projectile motion

Two-dimensional projectile motion refers to the motion of an object that is projected into the air at an angle to the horizontal, resulting in a parabolic trajectory. In the video, this concept is central as it discusses how an object moves in both the horizontal (X) and vertical (Y) directions independently. The script uses this concept to explain the problem-solving process for determining the horizontal distance an object travels when projected from a certain height with a given initial velocity.

💡Initial height

Initial height is the starting elevation from which the object is projected. In the script, the object is projected from an initial height of 45 meters, which affects the vertical motion and the time it takes for the object to hit the ground. This is crucial for calculating the time it takes for the object to move horizontally.

💡Initial velocity

Initial velocity is the speed at which an object starts moving. The script mentions an initial velocity of 37 m/s in the horizontal direction, which is key to determining the distance the object travels in the X direction. The video uses this velocity to calculate the object's motion in the absence of forces in the X direction.

💡Acceleration

Acceleration is the rate of change of velocity with respect to time. The video explains that in the X direction, there is no acceleration (0 m/s^2) because there are no forces acting on the object, while in the Y direction, the acceleration is due to gravity (-9.81 m/s^2), which is essential for calculating the time it takes for the object to fall.

💡Free fall

Free fall is the motion of an object under the sole influence of gravity. In the script, the object's motion in the Y direction is described as free fall, with the only force acting on it being gravity. This concept is used to explain the acceleration in the Y direction and how it affects the object's vertical motion.

💡Kinematic equations

Kinematic equations are formulas used to describe the motion of an object. The video uses these equations to solve for variables such as time, velocity, and position. The script specifically mentions using the kinematic equations to find the time it takes for the object to fall in the Y direction and then uses that time to find the distance traveled in the X direction.

💡Change in position

Change in position refers to the difference in position between the starting and ending points. In the video, the change in position in the Y direction is -45 meters, indicating the object falls 45 meters vertically. This value is used in the kinematic equations to solve for time.

💡Time

Time is a critical variable in kinematic equations and is used to relate motion to the passage of time. The script explains that the time it takes for the object to fall in the Y direction is the same as the time it takes to travel in the X direction, which is essential for solving the problem.

💡Horizontal distance

Horizontal distance is the linear extent of the object's travel in the X direction. The video aims to find this distance, which is calculated by multiplying the initial velocity in the X direction by the time it takes for the object to fall. The script uses this calculation to arrive at the final answer of 112 meters.

💡Forces

Forces are pushes or pulls upon an object that result in a change in its motion. The script explains that in the X direction, there are no forces acting on the object, resulting in zero acceleration, while in the Y direction, gravity is the only force acting on the object, causing it to accelerate downwards.

💡Parabolic path

A parabolic path is the curved trajectory that a projectile follows due to the effects of gravity while moving at a constant horizontal velocity. The video script describes the object's motion as following a parabolic path, emphasizing the separation of motion into horizontal and vertical components.

Highlights

The video discusses a problem involving two-dimensional projectile motion.

The goal is to determine how far an object moves in the X direction when projected from an initial height of 45 meters.

The object has an initial horizontal velocity of 37 m/s.

The object follows a parabolic path due to projectile motion.

In the X direction, there are no forces acting on the object, resulting in zero acceleration.

In the Y direction, the only force acting is gravity, causing a free-fall motion with an acceleration of -9.81 m/s^2.

The motion in the X and Y directions is independent but occurs over the same time period.

Kinematic equations are used to solve the problem, focusing on variables such as initial and final velocity, change in position, acceleration, and time.

The acceleration in the X direction is zero, and in the Y direction, it is -9.81 m/s^2.

The initial velocity in the Y direction is 0 m/s for free-fall motion.

The final velocity in the Y direction is not needed to solve the problem.

The change in position in the Y direction is -45 meters, indicating a downward movement.

The time taken for the object to fall in the Y direction is calculated to be 3.03 seconds.

The time for the object to move in the X direction is the same as in the Y direction.

The object travels a distance of 112 meters in the X direction.

The object lands 112 meters away from the launch point.

The entire motion takes 3.03 seconds.

The video provides a step-by-step guide to solving the problem using kinematic equations.

The importance of understanding the acceleration in both the X and Y directions is emphasized.

The video concludes with a call to action for viewers to subscribe, like, and comment.

Transcripts

okay in today's video I'm going to go

over a problem involving two-dimensional

projectile motion and in this video we

want to

determine how far an object moves in the

X Direction when it is projected from an

initial height of 45 M with an initial

velocity in the X direction that is in

the horizontal direction of uh has

initial velocity of 37

m/s and when that object is projected

and it leav leav this surface it's going

to follow this nice parabolic path and

as I said we would like to know how far

does the object travel in the X

Direction the change in position in the

X Direction now in order to do this

problem in order to understand this

problem there's a couple things you need

to know first of all you need to know

that the object is moving in the X

Direction and in the y direction and

it's doing those two things separate

from each other independent from each

other and we need to talk about for just

a moment what it's doing in the X and

what it's doing in the y direction in

the X

Direction the forces are balanced there

are no forces acting on the object in

the X Direction and because there are no

forces acting on the object in the X

Direction you need to know that the

acceleration in the X direction is zero

meters perss squared in the y direction

there is one force during this object's

path that's acting on this object

there's only one force and that is the

force of gravity and in the y direction

this object is really experiencing what

we call freef fall and therefore the

acceleration is equal to the

acceleration due to gravity on Earth

which is-

9.81 m/ second squared keep that in mind

two separate accelerations two separate

motions in the X and the y

direction now in order to solve this

problem we're going to have to use our

kinematic equations and therefore what I

like to do first is for the y direction

and also for the X Direction write down

all five of the variables that are

contained in the kinematic equations

initial and final

velocity change in position acceleration

and time I have that for the Y and for

the X direction we're going to fill what

we know what we don't know and what

we're looking for now we've already been

told the

acceleration in the X and the y

direction we know those things the

acceleration in the y direction - 9.81

m/s squ x Direction 0 m per second

squared the initial velocity in the y

direction for freef fall

motion is 0 m/s we were told the initial

velocity in the X direction is 37 m/s we

don't know and we don't need the final

velocity in this problem in the y

direction and remember the initial

velocity in the X direction is 37

there's no acceleration so if there's no

acceleration it's not changing its

velocity so therefore the final velocity

is also 37

m/s in the X Direction the change in

position in the y direction is minus 45

it starts here and it moves down

this direction is the negative so the

change in position is minus 45 don't

forget your negative signs we're looking

excuse me we're looking for the change

in position in the X Direction now

you'll notice here we have the time and

the time in the X and the y direction we

don't know either of those but another

important thing that you need to keep in

mind for this problem is that the time

in the X and the time in the Y are the

same remember this thing is moving

independently in the X and the y

direction the time it takes to travel

this portion of its path and the time it

takes to travel in the X direction are

the same we want to find find the change

in position in the X we need the time in

the X we can't find it with with the

information that we're giving but we can

solve for the change in time in the y

direction and they're equal so we're

going to solve for the change in time in

the y direction and then we're going to

use that time

as the time in the X Direction because

we know the time it takes to fall in the

Y and the time it takes to move

horizontally in the X are the same

remember that point it's important all

right so for now I'm going to get rid of

our values for the X because we're going

to solve for the time in the Y that

means we're going to get out our

kinematic equations we want to solve for

time once again you'll see we've been

given three values we're asked to solve

for a fourth each of our equations has

four four variables in it we need one

that has the time in it this one does

not so we know automatic we cannot use

that equation the other three equations

all have the time in them but in

addition to the time we have to know the

other three variables to solve for the

time or to be able to solve for the time

and this equation and this equation is

the final velocity we don't know the

final velocity therefore we cannot use

this equation and therefore we cannot

use this equation but we have this last

equation we want the time we know the

acceleration in the y

direction we know the initial velocity

and we know the change in position so

therefore we can use this equation that

says the change in position in the y

direction is equal to the initial

velocity and the time in the Y times the

time plus 12

a^2 okay now another thing we can do to

simplify this equation before we solve

is you'll notice the initial velocity is

zero that means the initial velocity

times the time is also equal to zero so

now we can this term goes to zero and

we're going to solve for the time now to

do that

algebraically I have to get the time by

itself not the time squared to get the

time by itself time equals I'm going to

multiply both sides by two

first that'll get rid of my my

1/2 I'm going to divide by the

acceleration and then I'm going to take

the square root of both sides and that

will give me that the time is equal to

the square root of two * the change in

position the Y divided by the

acceleration I multiplied both sides by

T by two I divided by a and then I took

the square root of both sides and that

gives me time equals this now I can

simply plug my values in time is equal

to the < TK of 2 * - 45 don't forget

your negative signs / - 9 .81 m/s

squared and you get the time that it

takes for the object to fall in the y

direction is

3.03 seconds all right now we said that

the time and the Y and the time in the X

are the same so I'm going to put down

that the time is 3.03 and the time in

the x is also

3.03 remember it's doing both of those

things at the same time

all right now you'll notice I know how

fast the object is going 37 m/ second

it's doing that for 3.03 seconds so I'm

going to Simply take that the distance

change in the x is equal to the speed or

in this case the velocity times the

time that is the distance is equal to 37

* 3.03 and that tells me that in the X

Direction the object travels 112 m

okay so there you go that is our final

answer for how far the object travels in

the X Direction 112 M so if we launch an

object from a height of 45 M with an

initial velocity in the horizontal

Direction in the X Direction 37 m/s the

object will travel a

distance of 112 m in the X direction

it'll land 112 m away okay and it takes

3.03 seconds to do that all right there

you go I hope you found that helpful

follow those simple steps remember that

the acceleration the y direction is

freef fall minus 9.81 remember the

acceleration in the X Direction because

the forces are balanced is zero remember

the time it takes for the Y and the time

it takes in the X are the same all right

and you can solve that problem all also

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