PHY 101 1nd Edition Lecture 8 Outline of Last Lecture I. 3.2 Components of a VectorII. Vector Addiction AlgebraicallyIII. 3.3 Displacement, Velocity, Acceleration in 2 Dimensionsa. Average Velocityb. Instantaneous Velocityc. Average Accelerationd. Instantaneous Acceleration Outline of Current Lecture IV. Projectile MotionV. Velocity ComponentsVI. DisplacementVII. Direction of VelocityCurrent LectureProjectile Motion- Two dimensional free fall motion- Motion can be decomposed in the x and y directions (horizontal and vertical); they are independent of each othero Initial velocity: vi = vox + voyo Along x-direction: constant speed motion vx = voxo Along y-direction: constant acceleration vy = voy = ayt- A simple way of understanding “decomposition” of a 2D motiono Imagine that you shine a parallel light beam on a projectile- Velocity Componentso Decomposition of the velocity into x and y directions Initial velocity vox = vcosΘ voy = vsinΘo Velocity at any time t vx = vcosΘ vy = vsinΘo We also know that Vx = vox (constant) Vy = voy + ayt + voy –gtThese 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.- G= -9.8m/s2- Displacemento Motion in each direction is reduced to a one D motiono Along the x direction, since ax = 0 it is a constant velocity motion Δx = voxt + (1/2)ayt2 = voyt – (1/2)gt2o This explains the parabolic path- Direction of velocityo The magnitude of the velocity: (Pythagorean theorem)o Initial: vo = (vox2 + voy2)1/2o At time t: v = (x2 + y2)1/2- Direction of the initial velocityo Θ = tan-1 (voy/vox)- Direction of the velocity at time to Θ = tan-1 (vy/vx)- Ex. o A rescue plane is at a constant v= 40m/s horizontally at height of 100m above theground the plane drops a package to a hikero a) where does the package land on the ground with respect to the position it wasreleased from? Draw coordinates when it hits the ground: y-displacement: Δy=-100m x-displacement: Δx=voxt Use equations for free fall objects to find t: Δy=voyt + ½ 9yt2 = ½ (-9.8.m/s2)t2 =-100 =-4.9t2 T2=100/4.9 t=4.5so Therefore: Δx=voxt =40(4.5) = 180m Tip: Use y to find time , then use time to find displacement.o B) What are the horizontal and vertical components of the package’s velocity right before it hits the ground? Solution: Horizontal: Vx = vox = 40 m/s Vertical: Vy = voy + at ay = -g Vy = -gt t = 4.5s Vy = (-9.8)(4.5) Vy = -44 m/so C) What is the packages speed just before it reaches the ground? Vx = vpx = 40m/s vy = -44 m/s V = (vx2 + vy2)1/2 V = (402 + (-44)2)1/2 59.5 m/so D) what is the impact angle? Θ= tan-1(vy/vx) =
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