PHSX 114, Wednesday September 3 Reading for this lecture: Chapter 3 (3-5 thru 3-6)Reading for this lecture: Chapter 3 (3-5 thru 3-6) Friday: Exam #1Friday: Exam #1 Reading for the next lecture (Mon.): Chapter 4 (4-1 thru 4-4)Reading for the next lecture (Mon.): Chapter 4 (4-1 thru 4-4) Homework for this lecture: Chapter 3, question: 11; problems: 16, 20, 30, 35Homework for this lecture: Chapter 3, question: 11; problems: 16, 20, 30, 35

Exam #1 is Friday (Sept. 5) Bring a #2 pencil, your ID, and a calculatorBring a #2 pencil, your ID, and a calculator Fill in name, six-digit ID number and code on scantronFill in name, six-digit ID number and code on scantron Twelve multiple choice questionsTwelve multiple choice questions Mix of conceptual and quantitative questionsMix of conceptual and quantitative questions Closed book, sheet of equations will be providedClosed book, sheet of equations will be provided See web site for equation sheet and study guideSee web site for equation sheet and study guide

Vector Subtraction -V has the same length, opposite direction from V-V has the same length, opposite direction from V A-B = A+(-B)A-B = A+(-B) exampleexample

Multiply a vector by a scalar 2V is a vector in the same direction, but twice as long as vector V2V is a vector in the same direction, but twice as long as vector V exampleexample

Projectile Motion (2-dimensional motion with constant downward acceleration) Motion under the Earth's gravityMotion under the Earth's gravity Assume ground is flat, no wind or air resistanceAssume ground is flat, no wind or air resistance Horizontal motion -- no acceleration, a x =0; velocity, v x, is constantHorizontal motion -- no acceleration, a x =0; velocity, v x, is constant Vertical motion -- acceleration, a y =-g=-9.80 m/s 2 ; velocity, v y, is changingVertical motion -- acceleration, a y =-g=-9.80 m/s 2 ; velocity, v y, is changing

Apply constant acceleration equations to projectile motion v x = v x0v x = v x0 v y = v y0 - gtv y = v y0 - gt x = x 0 + v x0 tx = x 0 + v x0 t y = y 0 + v y0 t - ½gt 2y = y 0 + v y0 t - ½gt 2

Components of the velocity v x0 =v 0 cos θ 0, v y0 =v 0 sin θ 0v x0 =v 0 cos θ 0, v y0 =v 0 sin θ 0 horizontal component of velocity is constant and equal to v x0horizontal component of velocity is constant and equal to v x0 vertical component of velocity equals v y0 at t=0, and changes throughout motionvertical component of velocity equals v y0 at t=0, and changes throughout motion v y is zero at the top of the arcv y is zero at the top of the arc

Vertical motion is independent of horizontal motion Demo

Vertical acceleration is the same for all objects (ignoring air resistance) DemoDemo

Solving projectile motion problems 1.Choose an origin for your x and y coordinates and make a sketch. 2.Write down your knowns and unknowns. 3.Write down separate equations for the horizontal and vertical motion. 4.Form a strategy for using your equations and your knowns to find what you are after. 5.Check that your answer makes sense and that units are OK. Example Problem

Your turn: Answer: 8.8 m x = v x0 t=(100m/s)t, when x= 50 m, t=0.5 s y = y 0 + v y0 t - ½gt 2 = (10 m) - ½gt 2 = (10 m) - ½(9.8 m/s 2 ) (0.5 s) 2 = 8.8 m An archer shoots an arrow at a wall that is a horizontal distance of 50 m away. The initial velocity of the arrow is 100 m/s in a horizontal direction. The arrow begins at a height of 10 m above the ground. How far above the ground is the arrow when it hits the wall?

Another example problem