Two Dimensional Motion (4 of 4) Horizontal Projection, Worked Example
Summary
TLDRIn 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
📐 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.
🕒 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.
Mindmap
Keywords
💡Two-dimensional projectile motion
💡Initial height
💡Initial velocity
💡Acceleration
💡Free fall
💡Kinematic equations
💡Change in position
💡Time
💡Horizontal distance
💡Forces
💡Parabolic path
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
thank you for watching I hope you found
that helpful if you did please do all of
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