good day students welcome back to ma estrang techie youtube channel we are now in quarter four of our grade nine science lesson and the first topic that we are going to have today is all about the horizontal and vertical motions of a projectile check out our learning objective describe the horizontal and vertical motions of a projectile get ready to learn this lesson so keep on watching [Music] in grade 8 you have learned about newton's second law of motion the law of acceleration according to these law the net force is equal to the product of the mass of the object and its acceleration let's have a simple activity to refresh your mind state the second law of motion by completing this paragraph so let us start the law of of course acceleration the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force in the same direction as the net force and inversely proportional to the mass of the object that is the second law of motion good job class it is very good that you can still recall the law of acceleration always remember that the larger the force applied on an object the larger its acceleration usually the acceleration of a body may change and vary from time to time due to several factors if that is the case when can we say that there is a constant acceleration let's have a review about uniformly accelerated motion or uam look at the picture shown in your screen are you familiar with these blue uniformed men who are usually positioned in the busy streets you might have seen them as you cross the streets to reach your school or on your way to the market or a friend's house they are the traffic enforcers who are they and what do they do they play a crucial role in maintaining the flow of traffic and in implementing loss and rules and regulations on the road that every motorist should follow they may also be considered as front liners amidst the coveted 19 pandemic as they assess soldiers and police officers in checkpoints speed limits are observed on roads they vary and depend on several conditions speed limits for highways are different from speed limits on small avenues and streets this prevents vehicles to speed up or accelerate unnecessarily in our bc streets as they are being monitored by traffic enforcers acceleration is defined as the rate of change in velocity with time it is a vector quantity having both magnitude and direction acceleration of vehicles usually changes from time to time a vehicle may speed up as it travels to highways or slow down as it approaches an intersection it should also help when the stop light is red and eventually accelerate when the light turns green we can say that most type of acceleration that we observe is not constant or non-uniform due to these factors when you are riding a vehicle you can say that speeds up or slows down based on its speedometer a speedometer indicates the speed of a vehicle it is usually combined with a device known as odometer but records the distance throttled assume that you are riding a vehicle and observe these speedometer readings what can you say about the readings on the speedometer during each minute what can you infer from the data in the table this means that vehicle speeds up every minute there is one kilometer per hour increase on the speed of the vehicle the increase for every minute is constant therefore we can say that the vehicle has uniform acceleration this vehicle exhibits a uniform or constant acceleration it is a type of motion in which the velocity of an object changes by an equal amount in every equal time period just like these example in uniformly accelerated motion or uam the value of the acceleration is constant it does not change the velocity changes but at a constant rate an object with zero acceleration is said to be in uniform motion an object in uniformly accelerated motion has non-zero but constant acceleration acceleration represents how velocity changes with time velocity represents how position changes with time next picture as you can see we have a coin toss it upward does it stay up no when it reaches its maximum height eventually it will go down what makes it go down the answer is gravity gravity pulls the coin down things that are thrown upward will go down because of these force gravity also pulls us towards the earth that is why we are not being thrown outside the space what floor was your classroom when you were in grade 7 in grade 8 was it on the 3rd or 4th floor which is easier climbing to your classroom or going down to the school canteen when you climb a high place you go against gravity when you are on a high place and you go down you are moving toward gravity so you use less force same is true with free falling objects objects at free fall also exhibit uniformly accelerated motion the acceleration due to gravity is 9.8 meter per second squared neglecting air resistance it is a constant acceleration for all falling bodies regardless their mass and weight all objects on the earth's surface are being accelerated towards the center of the earth at a rate of 9.8 meter per second squared this means that if you raise an object above the surface of the earth and drop it the object will start from res and its velocity will increase 9.8 meter per second for each second it is falling toward the earth's surface until it strikes the ground in general a uniformly accelerated motion is the one in which the acceleration of a body throughout the motion is uniform it can be observed in either vertical or horizontal dimension and into dimensions let us have an activity for you to understand the uniformly accelerated motion analyze each scenario identify if it exhibits uniformly accelerated motion or not let us have the first photo a bike at rest is this uam or not the answer not uam next a fruit dropping from a tree is this an example of uam or not the answer uam next a boy holding a book and we answer not uam next rocks falling from a cliff is this a uniformly accelerated motion or not the answer uam last picture a trok running with a constant acceleration is it uam or not the answer it is an example of uam great job class you did well now let us proceed let us take a look of these photo as you can see it is a photo of the man diving the motion of the man as it dives into the water is moving along a curved path due to a natural force called gravitational force or gravity gravity is the natural force that causes object to fall towards the earth hence the acceleration due to gravity is always directed downwards with this the motion of the mind undergoes projectile motion projectile motion is a form of motion where an object moves in a curved path the object that is thrown or projected and exhibits projectile motion is called projectile the curve-like path undertake by a projectile is called trajectory let us have another photo are you familiar with this game a sepak takraw player kicking a rotten ball over the net what is the projectile in that scenario if you think that the rattan ball is the projectile then you are right how do you describe its trajectory the rotten ball traveling on a curved path that is correct in mathematical terms we call this pattern parabola an angle launch projectile exhibits a full parabolic trajectory motion this shows that as the player kick the rattan ball in the air the ball will eventually go back to the ground still due to gravity as it moves horizontally hence projectile motion consists of horizontal and vertical motion working independently the two pictures and scenarios that i have shown you are the two types of projectile motion horizontally launched projectiles and angle launched projectiles let us take note of their components in horizontal motion it is the x component no resistance in any form to simplify we neglect air resistance next there is a constant horizontal velocity we use vx to represent it v for velocity and x for horizontal component next horizontal acceleration is zero why because there is no external force acting along the horizontal motion it means that acceleration does not exist and it is represented as a x a for acceleration and x for the x component next we call the horizontal distance as range and it is represented as dx d for distance and x for the horizontal component now let us go to the vertical motion this is the y component the force acting upon in this motion is the force of gravity and it is represented as g or a y or acceleration in the y or in the vertical motion or acceleration due to gravity the constant acceleration is always equal to negative 9.8 meter per second squared take note class that negative indicates the direction and the direction is downward due to the presence of gravity accelerating at 9.8 meter per second squared this causes the vertical velocity to increase which makes it not constant and it is represented as v y v for velocity and y for the vertical component or vertical motion next we call the vertical distance as the height and it is represented as h or d y for you to understand more let us have an example scenario this is an example of a horizontal launched projectile suppose you roll a marble on a frictionless table observe the motion of the marble as it goes down the floor the motion of the marble is gradually increasing from point a to point b due to the presence of gravity accelerating at 9.8 meter per second squared these cause the vertical velocity to increase which makes it not constant for the horizontal velocity there is no external force acting along the horizontal motion which means no acceleration resulting to a constant horizontal velocity let us have an example problem to solve a marble is thrown horizontally from a table top with a velocity of 1.50 meter per second the marble falls 0.70 meters away the table's edge find the following how high is the table and what is the final velocity of the marble just before it hits the cup first let us illustrate the problem a marble is thrown horizontally with an initial velocity of 1.50 meter per second we are going to represent it as vi or initial velocity next the marble falls at 0.70 meter away from the table therefore we can say that it is the horizontal distance or dx and that is again 0.70 meters next we are going to find how high the table is or our vertical distance or d y and also if we have initial velocity of course we have the final velocity of the marble before it touches the cup and it is represented as v f or the final velocity now let us solve the problem here are the given and the formulas that we are going to use again our given the horizontal distance is 0.70 meters our initial velocity which is equal to 1.50 meter per second and our acceleration due to gravity is 9.8 meter per second squared to find the height or the vertical distance of the table this is our formula get the one half of the product of acceleration due to gravity and the squared of time as you can see we do not have the value of time therefore we need to get first the time to get the height of the table so take note of this formula let us substitute our given to our formula time is equal to dx which is 0.70 meters divided by the initial velocity 1.50 meter per second let us divide 0.70 divided by 1.50 and the answer is 0.47 this is already rounded off next how about the unit as you can see we can cancel out meters so let's cancel out meters and the remaining unit is second therefore our final answer is 0.47 second now we can determine the height of the table let's substitute the given in our formula acceleration due to gravity times time which is 0.47 do not forget to get the square of the time divided by 2 squaring 0.47 we get 0.2209 seconds squared next let us simplify this by multiplying the acceleration due to gravity and time therefore we have 2.16482 divided by 2. as you can see we can cancel out second squared and the remaining unit is meters next let us divide this into two and the quotient is [Music] 1.08 meters this is also rounded off therefore this is our final answer the height of the table is 1.08 meters now let's proceed to our next question what is the final velocity of the marble just before it hits the cup as you can see we have an additional given that we got a while ago which is the height or the vertical distance 1.08 meters to determine the final velocity of the marble we need to get its resultant velocity or vr the resultant velocity of an object is the sum of its individual vector quantities remember that final velocity has two components the horizontal and the vertical so let us find the two velocity components and then solve for the resultant velocity using the pythagorean theorem equation to get the value of v fy this is the formula that we are going to use take note class that a g is similar to g so let us substitute the given to our formula to find the value of v f y 2 times acceleration due to gravity times the height multiplying these three quantities we have the product of 21.168 meters squared per second squared let us get the square root vfy is equal to negative 4.6 meter per second as you can see we also simplify the unit take note class that negative sign indicates downward direction so we can say that vfy is equal to 4.6 meter per second downward or negative 4.6 meter per second now that we have the value of vfy we can now determine the final velocity or the resultant velocity let us go to the next slide again this is our formula the faith algorithm theorem equation square root of v f y squared plus v f x squared we already have the value of v f y which is negative 4.6 meter per second now to get the value of the x squared take note class that initial velocity is also equal to the initial horizontal velocity and the final horizontal velocity which is also equal to the horizontal velocity therefore the value of v of x is 1.50 meter per second now let us substitute the given to our formula vr is equal to the square root of negative 4.6 meter per second squared plus 1.50 meter per second squared getting the squared of these two quantities we have 21.16 and 2.25 as you can see we also squared the unit at these two we have 23.41 meters squared per second squared get the square root and our final answer is 4.84 meter per second and this is our final answer and that is how you are going to solve this kind of problem take note class of these equations that may help you solve problems involving projectile launched horizontally and that ends our lesson this week check out the part 2 of this lesson before we end let us have a shout out shout out to shout out to kritom also shout out to marisol morphiel and shout out to bayawan city science and technology education center negros oriental thank you all so much for watching i really appreciate it if you like share subscribe and comment down below for a shout out see you guys on my next video bye [Music] you