Phys 201 1nd Edition Lecture 5 Outline of Last Lecture I. Solving for vectors and separating x and y equation components.Outline of Current Lecture II. General Projectile Motion EquationsIII. Solving Projectile Motion when Y=0A. Range EquationIV. Solving Projectile Motion Without Y DisplacementV. Solving Using VectorsCurrent LectureProjectile Motion EquationsProjectile motion refers to the movement of an object in a parabolic shape, being launched from a certain angle at a certain speed. An object launched in a projectile motion will typically move up and forward to a certain point, and then forward and down before stopping. Even though there is motion in both the x and y directions during projectile motion, there is not always both X and Y displacement. This means that to solve for variables in this kind of motion we separate our vector equations into x and y equations.X=Xo + VxotVx=VxoVx = Vxo + atVx2 = Vxo2 + 2(a)(x-xo)y=yo +Vyot -.5(g)t2 Vy=Vyo-gtVy2=Vyo2-2g(y-yo)Projectile Motion For No y Displacement:These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Initial velocity (Vo) tells us both the size and the angle of the motion as it has both magnitude and direction. If you shoot something off from the same height that it will land (i.e. if you throw something from the ground and it lands back on the ground), then both y and yo are equal to zero, causing the displacement in the y direction to also be zero. This means the unknown of your projectile motion will bethe change in the x position. To find this, you will need speed in x direction and time.The easiest way to find time is to solve using the equation; y=yo + Vyot - .5gt2Because we know y displacement is 0, this equation can be simplified to; t=2Vyo/gTo get Vyo, we will use the equation Vyo/Vo=Sin(θ), or Sin(θ)Vo=Vyo where θ= the angle of velocity and Vo=the velocity at which the object is launched. It is important that we use the equation involving Vyo to solve for time, as Vyo refers to the time in which the object is up in the air.Once we find our value for time, we can solve for x displacement by using the equation X=xo + Vxot where X is the displacement in the x position. We will solve for Vxo the same way we solved for Vyo, except you will change Sin to Cos, meaning your equation will be Vxo=VoCos(θ). Range Equation:We can find the range in how far an object can travel at a given speed and angle by using the equation; X=(2Vo2/g)(Cos(θ)Sin(θ)). Simplified, this equation reads; X=(2Vo2/g)Sin(2θ)Remember that this formula disregards y displacement. Therefore, this equation works ONLY if the y displacement of the object is 0.Projectile Motion with y Displacement:If y displacement is not 0, then we have to solve for projectile motion using both our x and y equations. Example: If I propel a steel ball with an initial speed of 6.5m/s the top of a 1m table at an angle of 30◦, how long will it take for the ball to hit the ground and how far will it go?First, we need to solve for t using the equation |t|= (Vy-Voy)/-gTo use this equation we will need to solve for Vy and Voy. First, We can find Voy using the equation Voy=VoSin(θ)Voy=6.5Sin(30)=3.25We can now solve for Vy using the equation Vy2=Vyo2-2g(y-yo)Vy2=3.252-2(9.81)(1)Vy2=30.6Vy=√30.6= 5.5m/sWhen we plug these values into our equation for time, we get |t|=(5.5-3.5)/-(9.81)=0.875sTo find out how far the ball traveled, we will plug this value for time into the equation X=Vxot. First we need to solve for Vox.Vox=VoCos(θ)=6.5Cos(30)=5.6X=(5.6)(.875)=4.9Projectile Motion and Vectors:Therefore, the ball travels 4.9 meters before landing.Projectile motion works a bit like back to back triangles. The triangle on the left represents the projectile motion of the object as it is going up. The triangle on the right represents the velocity when the object is on its way down. The hypotenuse of these two triangles corresponds to the objects velocity. If necessary, we can use what we know about vector x and y components and trigonometric functions to solve for the velocity of the object, as the x and y components represent the Vxo and Vyo, respectively.Example; If we want to catch the same ball from above in a tall, slim cylinder, at what angle would we have to tilt the cylinder so that the ball lands in it? What is the ball’s velocity just before it lands?If we imagine the ball’s downward motion as a triangle with x and y components, it would look like this;Vy=5.5Vx=5.6θWe can find the angle using the trigonometric function; Tan(θ)=opp/adjTan(θ)=5.5/5.6θ= Tan-1(5.5/5.6)=45◦To find the ball’s velocity, we can solve using our trigonometric formulas to solve for the hypotenuse, OR,we can use Pythagorean theorem and solve for V as;V=√(5.52+5.62)V=7.8m/s