Projectile Motion Launched at an Angle | Height and Range | Grade 9 Science Quarter 4 Week 2
Summary
TLDRThis engaging lesson from Maestrang Techy's YouTube channel dives into the intricacies of projectile motion launched at an angle. The video begins by revisiting the fundamental concepts of projectile motion, emphasizing the parabolic trajectory and the distinction between constant horizontal velocity and vertical acceleration due to gravity. The focus then shifts to the impact of the angle of release on a projectile's range and height. The channel illustrates that a 45-degree launch angle yields the maximum range, while a 75-degree angle results in the greatest height. The lesson is enriched with real-world examples, such as a baseball game, to contextualize the theory. Mathematical equations are introduced to solve for variables like maximum height and horizontal displacement, demonstrated through a practical problem involving a baseball hit at a 25-degree angle. The video concludes with a call to action, encouraging viewers to engage with the content and anticipate future lessons, making it an informative and interactive learning experience.
Takeaways
- đ The lesson focuses on the projectile motion launched at an angle, exploring the relationship between the angle of release and the height and range of the projectile.
- đ Projectile motion involves a parabolic trajectory with horizontal and vertical components; the horizontal component has constant velocity, while the vertical has constant acceleration due to gravity (9.8 m/sÂČ).
- đ The vertical velocity of a projectile changes throughout its trajectory: it decreases as the projectile rises due to gravity, becomes zero at the highest point, and increases as it falls back down due to the direction aligning with gravity.
- đ The game of baseball serves as an example of projectile motion launched at an angle, where the horizontal velocity (Vx) remains constant, and the vertical velocity (Vy) changes as described.
- đ When launching a projectile at an angle, the initial velocity can be resolved into horizontal and vertical components, with the horizontal velocity being constant and the vertical velocity affected by gravity.
- â±ïž The time it takes for a projectile to stop at its highest point is equal to the time it takes to return to the launch point.
- đą The initial velocity upward is the same magnitude as the final velocity when the projectile returns to its original height.
- đŻ The greatest range for a projectile is achieved when launched at a 45-degree angle, while the maximum height is achieved at a 75-degree angle.
- âïž Complementary angles, such as 30 and 60 degrees, result in the same range because they sum up to 90 degrees.
- đ As the angle of launch increases, the vertical displacement of the projectile also increases.
- đ At the highest point of its trajectory, the vertical component of the projectile's velocity is zero, and the time to reach this point is half of the total time of flight.
- đ An example problem is solved in the lesson, calculating the maximum height and horizontal displacement (range) of a baseball hit at an angle of 25 degrees with a velocity of 30 m/s.
Q & A
What is the basic concept of projectile motion?
-Projectile motion is the motion of an object thrown at an angle to the horizontal. It involves a parabolic trajectory with constant horizontal velocity and vertical motion with constant acceleration due to gravity.
What are the two components of projectile motion?
-The two components of projectile motion are the horizontal component, where the acceleration is zero and the velocity is constant, and the vertical component, where the acceleration is constant and equal to the acceleration due to gravity (9.8 m/s^2).
What is the relationship between the angle of release and the range of a projectile?
-The angle of release affects the range of a projectile. The maximum range is achieved when the projectile is launched at an angle of 45 degrees with respect to the horizontal.
What is the relationship between the angle of release and the maximum height of a projectile?
-The maximum height of a projectile is achieved when it is launched at an angle of 75 degrees. As the angle of launch increases, the vertical displacement of the projectile also increases.
What is the significance of complementary angles in projectile motion?
-Complementary angles, such as 30 and 60 degrees, are significant in projectile motion because they result in the same range. This is because the sum of the angles is 90 degrees, and the projectile's path is symmetrical around the vertical axis.
How is the vertical velocity of a projectile affected by gravity during its flight?
-The vertical velocity of a projectile decreases as it rises due to the opposing force of gravity. At the highest point, the vertical velocity is zero as the projectile momentarily stops before descending. As it falls back to the ground, the vertical velocity increases because the direction of motion aligns with the gravitational force.
What is the formula used to calculate the maximum height reached by a projectile?
-The formula to calculate the maximum height (dy) reached by a projectile is given by: dy = (vi * sin(theta))^2 / (2 * g), where vi is the initial velocity, theta is the angle of launch, and g is the acceleration due to gravity.
How do you calculate the horizontal displacement or range of a projectile?
-The horizontal displacement (dx) or range of a projectile is calculated using the formula: dx = vi * cos(theta) * t, where vi is the initial velocity, theta is the angle of launch, and t is the total time of flight.
What is the total time of flight for a projectile launched at an angle?
-The total time of flight for a projectile is calculated using the formula: t = (2 * vi * sin(theta)) / g, where vi is the initial velocity, theta is the angle of launch, and g is the acceleration due to gravity.
What is the significance of the initial velocity in projectile motion?
-The initial velocity is significant as it determines the magnitude of both the horizontal and vertical components of the projectile's motion. It affects the range, maximum height, and the overall trajectory of the projectile.
How does the time taken to reach the maximum height relate to the total time of flight for a projectile?
-The time taken to reach the maximum height is half of the total time of flight. This is because the projectile takes an equal amount of time to ascend to its highest point and to descend back to the point of launch.
Outlines
đ Introduction to Projectile Motion at an Angle
This paragraph introduces the topic of projectile motion launched at an angle, emphasizing the importance of understanding the relationship between the angle of release and the projectile's range and height. It recaps the basic concepts of projectile motion, such as trajectory and the distinction between horizontal and vertical components. The horizontal component maintains a constant velocity with zero acceleration, whereas the vertical component experiences constant acceleration due to gravity (9.8 m/sÂČ). The concept is illustrated using the example of a baseball, highlighting how the vertical velocity changes as the projectile rises, reaches its peak, and falls back down. The variables and facts involved in projectile launch at an angle are also discussed, along with the equations that can be used to solve related problems.
đŻ Optimal Angles for Range and Height in Projectile Motion
The second paragraph delves into the optimal angles for achieving maximum range and height in projectile motion. It reveals that a 45-degree angle results in the greatest range, while a 75-degree angle leads to the maximum height. The concept of complementary angles is introduced, explaining that a projectile launched at 30 degrees will have the same range as one launched at 60 degrees because these angles sum up to 90 degrees. The paragraph also discusses how the vertical displacement and the time to reach the maximum height change with the angle of launch. An example problem is presented, where a baseball player hits a home run, and the task is to find the maximum height and horizontal displacement of the ball, using given values for the angle and velocity.
â Solving for Maximum Height and Range in Projectile Motion
In the final paragraph, the video script provides a step-by-step solution to the example problem introduced in the previous section. The process involves calculating the maximum height reached by the baseball using the initial velocity, angle, and acceleration due to gravity. The formula used is vi * sin(theta)ÂČ / (2 * g), which, after substitution and calculation, yields a maximum height of 8.20 meters. Subsequently, the horizontal displacement or range is determined using the formula vi * cos(theta) * time, where the total time of flight is calculated first. After solving for time and applying it to find the horizontal displacement, the result is 70.42 meters. The paragraph concludes with a summary of the lesson on projectile motion and an invitation for viewers to engage with the content by liking, sharing, and subscribing for future lessons.
Mindmap
Keywords
đĄProjectile Motion
đĄTrajectory
đĄAcceleration
đĄAngle of Release
đĄRange
đĄMaximum Height
đĄHorizontal Velocity (Vx)
đĄVertical Velocity (Vy)
đĄComplementary Angles
đĄTime of Flight
đĄInitial Velocity
Highlights
The lesson focuses on the relationship between the angle of release and the height and range of a projectile in projectile motion.
Projectile motion is characterized by a parabolic trajectory with constant horizontal velocity and constant acceleration in the vertical direction due to gravity.
The horizontal component of projectile motion has zero acceleration, maintaining a constant velocity.
The vertical component of projectile motion is influenced by gravity, causing a constant acceleration of 9.8 m/sÂČ.
Baseball is used as an example of projectile motion launched at an angle, illustrating the changes in vertical velocity due to gravity.
The vertical velocity of a projectile decreases as it rises, stops momentarily at the highest point, and increases as it falls back down.
The initial velocity of a projectile launched at an angle can be resolved into horizontal and vertical components.
The time taken for a projectile to stop at its highest point is equal to the time it takes to return to the launch point.
The initial velocity of a projectile is equal in magnitude to the final velocity when it returns to its original height.
The greatest range in projectile motion is achieved at a 45-degree angle.
The maximum height is achieved at a 75-degree angle launch.
Complementary angles, such as 30 and 60 degrees, result in the same range for a projectile.
As the angle of launch increases, the vertical displacement of the projectile also increases.
The vertical component of velocity is zero at the highest point of a projectile's trajectory.
The time to reach the maximum height is half of the total time of flight for a projectile.
An example problem involves calculating the maximum height and horizontal displacement of a baseball hit at a 25-degree angle with a velocity of 30 m/s.
The formula for calculating the maximum height involves the initial velocity, sine of the angle, and acceleration due to gravity.
The horizontal displacement or range is calculated using the initial velocity, cosine of the angle, and the time of flight.
The lesson concludes with a practical application problem and a summary of the key concepts in projectile motion.
Transcripts
good day students welcome back to maestrang techy youtube channel let us continue our discussion Â
if you haven't watched our week one video lesson about the horizontal and vertical Â
motions of a projectile check out the link in the description box below we are now going to have Â
grade 9 science quarter 4 week 2 lesson which is all about projectile motion launched at an Â
angle here's our learning objective investigate the relationship between the angle of release Â
and the height and range of the projectile so get ready to learn this lesson and keep on watching
from the previous lesson you are introduced to the basic concepts of projectile motion such as Â
trajectory and the definition of projectile motion itself a body in projectile motion has Â
been established to have a parabolic trajectory with a horizontal and vertical components the Â
horizontal component of a projectile motion has the acceleration equal to zero since the velocity Â
is constant on the other hand the vertical component of acceleration is constant which Â
is acceleration due to gravity and that is always equal to 9.8 meter per second squared therefore Â
projectile motion is the combination of horizontal motion with constant velocity and vertical motion Â
with constant acceleration take a look at this are you familiar with this game yes baseball Â
this is an example of projectile motion launched at an angle for angle launch projectile horizontal Â
velocity or vx is still constant while the vertical velocity can be described in three Â
parts first from the picture as you observed the projectile rises from point a to point b Â
the vertical velocity or v y is decreasing this is because the direction of gravity is opposite Â
to the projectile motion next as the projectile reaches the maximum height which is the point b Â
it momentarily stops causing a vertical velocity or v y equal to zero and third when it returns Â
back to the ground from point b to point c it agrees to the direction of gravitational force Â
hence vertical velocity is increasing so when the vertical velocity of the baseball as it rises to Â
the air decreases due to the opposing direction of gravity towards the motion when the baseball Â
reaches the maximum height it momentarily stops causing the vertical velocity to be zero when Â
it reaches to the ground its vertical velocity increases since the direction of the baseball's Â
motion is the same with gravity take note of that class now take a look at the variables involved in Â
projectile launch at an angle we have here the horizontal component and the vertical component
next we have the facts about projectile launched at an angle Â
first up an object is projected from rest at an upward angle theta just like this scenario Â
the ball started from rest where stephen carey is holding the ball Â
second its initial velocity can be resolved into two components as you can see we have the Â
horizontal and the vertical component third the horizontal velocity is constant due to gravity Â
a constant horizontal velocity that moves in the same direction as the launch the acceleration of Â
which is zero fourth the amount of time the object takes to come to a stop at its highest point is Â
the same amount of time it takes to return to where it was launched from and lastly the initial Â
velocity upward will be the same magnitude as the final velocity when it returns to its original Â
height so these are the facts about projectile launched at an angle next here are some of the Â
equations that may help you solve problems involving projector launched at an angle
let's proceed take a look at this photo class what can you say which angle results Â
in the greatest range when we say range it is the horizontal displacement Â
and as you can see the farthest range is in the 45 degrees angle next question
which angle results in the maximum height
as you can see it is the 75 degrees angle how would you compare the distance traveled Â
by projectile launch at 30 and 60 as you can see they have the same range same as 15 and 75 Â
they have the same range this scenario that i have shown you is also an example of projectile motion Â
launched at an angle and these are the possible results if you launch an object at different angle Â
take note class angle that is usually represented by theta is a numerical value in degrees Â
expressing the orientation of a projectile to be thrown to sum it up class the angle of release Â
affects the range and height of a projectile the maximum range is achieved if the projectile is Â
fired at an angle of 45 degrees with respect to the horizontal component Â
an object launched at an angle of 30 degrees will also be the same if it is launched at 60 degrees Â
the angles 30 and 60 degrees are called complementary angles because they add up Â
to 90 degrees as the angle of launch increases the vertical displacement of the projectile will also Â
increase at the highest point the vertical component of velocity is zero and the time Â
to reach the maximum height is half of the total time of flight now let us have an example problem Â
a baseball player leads off the game and hits a long home run the ball leaves the bat at an Â
angle of 25 degrees with a velocity of 30 meter per second let us find the maximum height reached Â
by the ball and the horizontal displacement of the ball let us illustrate the problem as you can see Â
we have an angle of 25 degree and a velocity of 30 meter per second Â
we are looking for the maximum height reached by the ball and the horizontal displacement or range Â
or dx of the ball let us try to solve this problem here are the given our initial velocity Â
or vi which is equal to 30 meter per second our degree of angle which is 25 degrees acceleration Â
due to gravity which is 9.8 meter per second squared the formula that we are going to use is Â
v i times sine theta squared divided by twice the acceleration due to gravity Â
now let's substitute the given to our formula d y is equal to our vi which is 30 meter per second Â
and sine theta which is sine 25 degrees do not forget to square it to itself divided by Â
2 times 9.8 meter per second squared multiplying these two quantities and squaring it we have the Â
product of 160.745 and so on meters squared per second squared divided by the product of Â
2 and 9.8 and that is 19.6 meter per second squared let us divide this two the quotient Â
8.20 and as you can see we have to simplify the units let's cancel out and the remaining unit Â
is meter therefore our final answer or the maximum height reached by the ball is 8.20 Â
meters now let us solve the second one what is the horizontal displacement or range of the ball Â
again here are our given the formula that we are going to use to find the dx Â
is just multiplying v i cos sine theta and the time as you can see we do not have the value of Â
time therefore we have to solve the total time to proceed in the x and this is the formula that Â
we are going to use so let us solve the total time is equal to 2 times our vi and sine theta all over Â
the acceleration due to gravity 30 times sine 25 degrees is equal to 12.678 meter per second Â
divided by of course our acceleration due to gravity 12.678 divided by 9.8 times 2 we have Â
2.59 let us not forget to simplify our unit by canceling and our unit is seconds therefore Â
the total time traveled by the ball is 2.59 seconds now we can now solve for the value of Â
dx dx is equal to our vi cosine theta and the value of time multiplying these free quantities Â
our final answer is 70.42 let us not forget to cancel the units therefore our final answer is Â
70.42 meters and that ends our lesson about projectile motion i hope you learned something new Â
today please give this video a thumbs up share this to your classmates and do not forget to Â
subscribe to keep you updated for our next video lesson comment down for a shout out shout out to Â
gabriel balitos al qaeda nifty red gian and mamirina victas and all the grade 9 students of Â
san bartolome high school and also shout out to all the science teachers of the Â
vitae national high school thank you all so much for watching see you on my next video bye you
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