TwoDimensional Motion Vertical Horizontal Vertical Motion Galileo was

Two-Dimensional Motion Vertical & Horizontal

Vertical Motion Galileo was the first to show that all objects fall to Earth with a constant acceleration. Acceleration due to gravity (g) is the change in velocity over a period of time in a downward vertical direction. Value of “g” = 9. 80 m/s 2

Directions and Signs A. Up v, t, & d are (+) g is (-) B. Down v, d, & g are (-) t is (+) “g” is negative regardless of direction. “t” is always positive

Equations V 2 = V 1 + gt V 22 = V 12 + 2 gd d = V 1 t + ½ (gt 2)

Free-Fall I An object dropped from rest takes 4 s to reach the ground. What is the final velocity of the object? » m/s g(-) d(-) v(-) t (+) v 1 = 0 NO! » » » v 2 = v 1 + gt 0 m/s + (-9. 8 m/s 2) (4 s) v 2 = - 40 m/s

Free-Fall II An object is thrown straight upward with a velocity of 10 m/s. What is the maximum height attained? V 2 = 0 m/s V 1 = +10 m/s g = -9. 80 m/s 2 V 22 = V 12 + 2 gd wee

Solution: Free –Fall II d = V 2 2 – V 12 2 g = (0 m/s)2 - (10 m/s)2 2 (-9. 80 m/s 2) d= 5 m

Symmetry of Fall Time up = time down Distance up = distance down Velocity up = velocity down

Testing Your Reaction Time Ask a friend to hold a ruler just even with the top of your fingers. Then have your friend drop the ruler. Taking the number of centimeters the ruler falls before you can catch it, calculate your reaction time. Average three trials.

Reaction Time Results The reaction time for most people is more than 0. 15 s.

Blue Physics Textbook Read pages 76 – 79 Solve the problems listed below in your Classwork/Homework Notebook p. 77: 25 p. 79: 27 - 30

Two-Dimensional Motion II Object rolled horizontally Components Vx = value of vector VY = 0 m/s Displacement is a vector quantity. d = V xt velocity Vx = horizontal

Horizontal Velocity & Displacement If the object is moving with constant velocity, the acceleration (horizontally) is zero. velocity acceleration Vx at beginning is equal to Vx at any point.

Path is called a trajectory Vx = 10 m/s at all points along the path d x = V xt

Vertical Velocity Vertical velocity increases with time until the object reaches terminal velocity. Terminal velocity - Velocity is no longer changing.

Vertical Displacement Vertical displacement increases with time. If d = V 1 t + ½(gt 2) and V 1 = 0 then d = 1/2 gt 2 = 4. 9 m/s 2 (t 2)

Time (s) Displacement (m) 0 0 1 4. 9 2 19. 6 3 44. 1

Graphs of Motion Displacement is increasing with time. Velocity is changing with time. Object is accelerating in the “y” direction

Horizontal & vertical motions are independent of each other Each type has its own equations Both the X- and Y- components must be used to determine d 2 and v 2 at any point. ** Time down is equal to time across at all points. **Back to the RIGHT TRIANGLE

X Y EQUATIONS d. X = v. Xt Use these equations for any point on the trajectory d = v 1 t + 1/2 gt 2 v 2 = v 1 + gt v 22 = v 12 + 2 gd v 2 =√ vx 2 + vy 2 tan Θ = vy/vx

EXAMPLE A car runs off a 43. 9 m cliff and lands 87. 7 m from the base of the cliff. a) How long does it take the car to reach the ground? b) What was the initial velocity of the car just before it left the cliff?

t 2= m) 2 (-43. 9 -9. 80 m/s 2 t = 2. 99 s d x= v xt vx = d/t = 87. 7 m 2. 99 s vx = 29. 3 m/s

Velocity Upon Impact With Ground Vx = 29. 3 m/s Vy 2 = vy 1 + gt = 0 m/s + (-9. 80 m/s 2)(2. 99 s) Vy = 29. 3 m/s V f 2 = v x 2 + v y 2 = (29. 3 m/s)2 + (29. 3 m/s)2 Vf = 41. 4 m/s

Object Projected at an Angle An object is shot into the air at a speed of 20. m/s and an angle of 40. o.

Things You Know Time up = time down Acceleration in the vertical direction is always 9. 80 m/s 2. Acceleration in the horizontal direction is always 0 m/s 2. Velocity at the top of the flight is 0 m/s.

Velocity vector is acting at an angle Use cos to determine the value of the Vx component and sin for the Vy component. Vx doesn’t change throughout trip. Velocity at release point = velocity at endpoint.

QUESTIONS AND SOLUTIONS Step I: Calculate the x- and ycomponents Vx = V cos Θ = 20 m/s cos 40 o Answer: 15 m/s Vy = V sin Θ = 20 m/s sin 40 o Answer: 13 m/s

Step II: Calculate the time in air t = vy 2 – vy 1 g = -13 m/s – (+13 m/s) -9. 80 m/s 2 t = 2. 7 s

Calculate the Range (where the object will land) d = V xt = (15 m/s)(2. 7 s) d = 41 m

Where is the object 0. 5 s after being launched? Step I: Calculate d in the y-direction d = V 1 t + ½ at 2 = 13 m/s (0. 5 s) + ½(9. 8 m/s 2)(0. 5 s)2 dy = 5. 3 m

Step II: Calculate d in the x-direction dx = vx t = (15 m/s) (0. 5 s) dx = 7. 5 m

Step III: Calculate the Resultant Displacement df 2 = dx 2 + dy 2 = (7. 5 m)2 + (5. 3 m)2 df = 9. 2 m

Calculate the Angle tan Θ = dy/dx = 5. 3 m ÷ 7. 5 m = 0. 688 = 35 o