okay good morning today let's have a look at chapter one lesson three and today we're going to have a look at the first half of this lesson which is about projectile motion and next lesson we'll have a look at the relative motion okay first of all let's have a look at the concept of projectile motion in sports in which a player kicks throws or hits a ball across the field or Court the player's initial contact with the ball proposable Upward at a angle the ball Rises to a certain point and gravity eventually curves the path of the ball downward if you ignore the effects of air resistance and Earth rotation the curved path or trajectory of the ball under the influence of Earth's gravity it follows the curve of a problem as shown here in figure one the ball acts like a projectile which is an object that is moving through the air and accelerating due to gravity the X direction is horizontal and positive to the right and the y direction is vertical and positive upward so here's the ball in figure one was hit with the tennis bracket if you draw imaginary lines through the four images you can trace the problem from where the ball made contact with the racket till the other end another imaginary line shows the uppermost point of the trajectory which is at the top of the high school after the ball leaves a rocket its past curves upward to this highest point and then curves downward you can see the symmetry of the balls pass because the shape of the upper bound curve exactly matches the shape of the downward bound cave anyone who has tossed any kind of object into the ear has observed this problem roblotic trajectory called projection motions before we formally Define projection motion projectile motion we will look at its properties first okay so suppose you drop a soccer ball from the roof of one story building well your front stands next to you and kicks another soccer ball horizontally at the same instant well they both land at the same time some people are surprised to learn that the answer is yes so here as shown here in figure two this figure shows a strong image of two balls released simultaneously one with horizontal projection as your friends soccer ball hat the horizontal lines represent equal time intervals which is a time interval between the camera's stroke flashes is constant the vertical components of the displacement increase by the same amount of each Pole the horizontal displacement of the projectile in this leak sense the ball is called the horizontal range which is usually denoted by Delta DX the horizontal motion is also constant subtract the trajectory forms from the combination of the independent horizontal and vertical motion we observe the following properties about the motion of a projectile first the horizontal motion of a projectile is constant then the horizontal component of acceleration of a projectile is zero next the vertical acceleration of a projectile is constant because of gravity and last the horizontal and vertical motions of a projectile are independent but they share the same time so combining these properties can help us Define projectile motion so by definition projectile motion is the motion of an object such that the horizontal component of the Velocity is constant and the vertical motion has a constant acceleration due to gravity the most important property of projectile motion in two Dimensions is that the horizontal and vertical motions are completely independent of each other this means that motion in One Direction has no effect on Motion in the other direction this allows us to separate a complex two-dimensional projectile motion problem into two separate simple problems one that involves horizontal uniform motion and one that involves vertical uniform acceleration down so here as shown in figure 3 this figure shows a basketball player hitting a Fly ball and pass it follows you can see that the horizontal velocity v x is independent of the vertical velocity v y you can also see this result in the stroke images in figure 2. the ball on the left simply drops but the ball on the right has an initial horizontal velocity the ball on the left Falls straight down while the bottom right follows a probiotic past typical of a projectile motion the balls have quite different horizontal velocities at each flash point in the image lastly they are at an identical Heights at each point this shows that their displacements and velocities along the Y Direction are the same the image confirms that the motion along the vertical direction does not depend on the motion along the horizontal Direction okay so now let's have a look at how we can analyze the projectile motion so in previous sections we have reviewed the equations that describe Motion in one dimension right you can use this same equations to analyze the motion of a projectile in two dimensions we can simply apply the equation to the to the X and Y motions separately because for each Direction that's simply a motion in one dimension right assumes that t equals zero the projectile leaves the origin and a initial velocity v i if the velocity Vector makes an angle Theta with the horizontal where Theta is a project projection angle then from the definitions of the cosine and sine functions we have the following relationship okay first of all the horizontal component v i x which is a horizontal component of the initial velocity is v i times cosine Theta foreign component in the vertical Direction equals v i times sine Theta where v i x is initial velocity which is at T equals 0 in the X Direction and v i y is initial velocity in the y direction also note that these numbers should have either positive or negative symbols in front of them negative symbols means the direction is to the opposite of your specified positive direction okay so here table 1 summarizes the kinetics kinematics equations we can use with both horizontal and vertical components so in horizontal motion X we can have a constant velocity equation for the X component which means v i x equals v i cosine Theta and v i s v i x is constant also because it's a constant velocity motion then we have the displacement in the X Direction equals v i x times delta T and put all values in you can have Delta DX equals v i times cosine Theta times delta T and along the vertical motion which is in the y direction we know that's a constant acceleration motion right the constant acceleration has a magnitude of about 9.8 meters per square second near the surface of Earth then we have a swallowing relationship v f y equals v i sine Theta minus G delta T Delta d y equals v i sine Theta times delta T minus 1 over 2 G times squared delta T and squared vfy equals v i sine Theta Square minus G times Delta d y also you should be careful with the positive or negative symbol in front of each quantity here okay if the direction of a quantity is in accordance with the positive direction you specified the number should be positive otherwise that should be a negative number okay so now let's have a look at some point sample question to illustrate these Concepts well let's consider this question here airplane carries release supplies to a multiple stranded in a snowstorm the pilot cannot safely land so he has to drop the package of supplies as he flies horizontally at a height of 350 meters over the highways the speed of the airplane is a constant which is at 52 meters per second figure 4 shows packages as it leaves the airplane or in meat shop or when it lands on heavy so there are two questions first calculate how long it takes for the package to reach the highway then determine the range of the package okay I'll give you five minutes to consider this problem by yourself and then we'll have a look at the solution together thank you okay have you finished let's have a look at the solution together so in question a we are given Delta d y which is negative three five zero meters and initial velocity is 52 meters per second and we want to calculate delta T right so we can set the I which is the initial velocity equals zero as the attitude at which the plane is flying therefore Delta d y is negative 350 meters so we can calculate the altity from the formula for the displacement along the Y Direction Delta d y equals v i times sine Theta delta T minus 1 over 2 times G times squared delta T okay so following this concept we can find Delta y Delta d y equals negative 1 over 2 times G times squared delta T and from this equation we can find delta T is square root of negative 2 times Delta d y over G and then put all values in we have delta T equals 8.45 seconds so the package takes about 8.5 seconds to reach the highway any questions Okay so for the second question we are given again Delta d y equals negative 3 15 meters v i equals 52 meters per second and we have already calculated that delta T equals 8.45 seconds right so this time we require Delta DX which is the range of the package so we can calculate Delta DX using the definition of cosine Delta DX equals v i times cosine Theta times delta T foreign Delta DX equals 52 times cosine 0 degrees times 8.45 and this gives us 440 meters so the range of the package well if you use scientific notation that should be 4.4 times 10 raised to the power of 2 meters okay so have a look at the solution here you have any questions foreign don't have any questions let's consider another practice question okay so this time we have a golfer who hits a golf ball with a initial velocity of 25 meters per second at a angle of 30 degrees so this time we have an angle of 30 degrees here above the horizontal which is Illustrated in figure 5 here scoffer is at initial height of 14 meters above the point where the ball lands then can you calculate the maximum height of the bowl and then determine the pulse velocity on Landing okay again I'll give you five minutes consider this question by yourself and then we'll have a look at solution together okay foreign minutes okay yes is foreign okay so let's have a look at a solution of this problem so first of all you have to choose a reference point right so suppose we choose the golfer's initial position as a reference point then the maximum height of the ball is the highest height of the ball with respect to the golfer at its initial position so we are given the initial velocity of the ball is VI equals 25 meters per second and the angle that the initial velocity makes is the positive horizontal Direction is 30 degrees and it requires the Delta d y Max so when the Gulf ball reaches its maximum height the Y component of the Bose velocity is zero right so actually there are a number of ways we can solve this you could either solve the time taken for the ball to reach the highest point first and then find the heist highest Delta d or you can also use VF Square minus v i square equals 2gs 2G Delta V right okay suppose we in this question we are going to find the time taken to reach the highest point first okay so we have the following relationship Delta d y equals v i times sine Theta times delta T minus 1 over 2 times G times delta T square right so in on the other hand we have vfy equals v i sine Theta minus G delta T and put all values in we have delta T equals v i sine Theta over G this gives us 1.28 seconds and then we can find Delta d y Max equals v i sine Theta delta T minus 1 over 2 G times Delta t squared then you have all the numbers ready there so plug these values in we have Delta d y Max equals 8 meters so the maximum height of the ball is 8 meters with respect to the initial position of the golfer okay then let's consider question B in question B work even again initial velocity and also we have calculated uh and we also we have known that Delta d y is 14 meters so we can if we set TI which is the initial velocity initial displacement at the point at which the golfer struck the golf ball then we have Delta d y equals minus 40 meters right so it all depends on which reference point you choose then use the equation vfy squared equals v i y squared minus 2G Delta y to calculate final vertical velocity of the ball before it hits the ground then calculate velocity when the boat lands using VF equals the square root of squared V of x squared plus v f y square and the inverse tangent ratio okay so following this logic we have vfi square equals v i y squared minus 2G Delta d y which equals v i signs Theta squared minus 2G Delta d y so put all the values in then we have vfy is either positive or negative 20.8 meters per second we have to choose a negative number because the object is moving down which is opposite to the positive direction along the y-axis you choose right then VX equals v i times cosine Theta which is 25 times cosine is 30 degrees which is 21.7 meters per second then we know final velocity VF is a square root of VX squared plus v y square and this gives us 30.1 meters per second so this is the magnitude of the final velocity and the angle that this velocity makes with the horizontal foreign direction is arc tangent 20.8 over 21.7 which is about 44 degrees so the final statement is the velocity of the ball one lens is 30.1 meters per second with Direction 44 degrees below the horizontal okay have a look at this question if you have any questions to ask thank you if no questions let's have a look at another concept which is the range equation for the projectile motion so when you know the initial velocity and launch angle of a projectile we can then calculate the projectile's range which is usually denoted by Delta DX foreign tile that lands at staying fine it started up it started from as shown here in this figure bigger six in this case Delta d y equals zero right because the total displacement is zero and we can a total displacement in the y direction is zero right because it lands at same height it started run and we can use this back to significantly simplify the equations of motion you can calculate the range using the equation Delta DX equals v i x times delta T if we know the initial velocity and the launch angle so to determine the value of delta T we can use the equation for vertical motion v i y which is a initial velocities component along the y direction equals v i times sine Theta the outer DUI equals v i sine Theta times delta T minus 1 over 2 times G times delta T Square the final level is the same as initial level so Delta d y equals zero right substituting values in the vertical motion equation gives us zero equals v i sine Theta times delta T minus 1 over 2 times G times Delta t squared well put the common factor delta T in front we have 0 equals delta T times V sine v i sine Theta minus 1 over 2 G delta T therefore we have either delta T equals zero on takeoff or v i times sine Theta minus 1 over 2 G delta T equals zero on Landing solving this later equation for delta T gives us the following equation delta T equals 2 v i sine Theta divided by G now we return to the equation for range the range Delta DX equals initial velocity v i in the as a component of the initial velocity along the x axis times delta T right substituting delta T and the initial velocity in the X direction we have vix equals v i cosine Theta this gives the Delta DX equals v i x times delta T which is v i times cosine Theta times 2 VI sine Theta over G and simplify this gives us Delta DX equals v i Square over G times 2 sine Theta cosine Theta also note that in mathematics we know two sine Theta cosine Theta is sine 2 Theta right this is according to the double angle formula so the final solution is here this is a range of such a projectile motion the range of this projectile is Delta at DX is given by v i Square over G times sine 2 Theta where v i is the magnitude of the initial velocity of a projectile launch at a and go Theta to the horizontal note that this equation applies only one Delta d y equals zero that is only one projectile lens at the same height from which it was launched the largest value of the range is one sine Theta sine two theta equals one because the sine function has a maximum value of 1 right so this maximum value occurs once the angle is 90 degrees since 2 theta equals 90 degrees then Theta must be 45 degrees so we know the largest value of the range can't occur once C High equals 45 degrees right so all the previous discussion and examples of projectile motion have assumed the air resistance is negligible this is close to the true situation in cases involving relative dense objects moving at low speeds such as shock use in short foot in as such as shock used in short foot combination however for many situations you cannot ignore the air resistance and you consider air resistance the analysis of project motion becomes more complex and it's beyond the scope of this context here so let's have a look at a sample problem in which we use kinematics equations to calculate the maximum height and the range for a projectile so now let's consider this question here suppose you kick a soccer ball at 28 meters per second to world to go at a launch angle of 21 degrees two questions you need to answer first how long does the soccer ball stay in air then determine the distance the soccer ball would need to cover to score a goal which means need to find the range of the soccer ball okay again I'll give you five minutes just think about this question by yourself first all right let's have a look at some different questions okay okay question a we are given initial velocity is 28 meters per second and the direction Theta is 21 degrees which is uh the last it makes with the Positive horizontal Direction and we need to find delta T okay so according to the range function and range formula we have delta T equals till we are sine Theta over G right then we know v i we know three thousand put all values in delta T equals 2.0 seconds so the soccer ball will stay in the air for 2.0 seconds oh sorry not that not the range formula that's the formula for delta T in second question we need to use the formula for the range right the range of the soccer ball is Delta DX is v i Square over G times sine 2 Theta and we know VI we know G and also we know Theta so put all the values in the solution is Delta DX is 54 meters so the range of the soccer ball is 54 meters okay any questions okay if you don't have any questions that's the end of today's class remember to complete your homework in Google classroom and submission before the line okay I'll see you next time