Motion in Two Dimensions Holt Physics Pages 98 – 106

Things to Remember one-dimensional motion was either all in the horizontal (x) direction or all in the vertical (y) direction acceleration of gravity (ag or g) is -9.8 m/s2 in the downward direction gravity acts the SAME on ALL objects!! downward velocity or position represented by “-“ find the components of any vector the same way as previous section

Also remember…

Equations Recall from chapter 2 when we studied 1-dimensional motion: v = Δx/Δt (used with constant velocity) Δy = vΔt + ½ aΔt2 vf2 = vi2 + 2aΔy c2 = a2 + b2

Reviewing the Math #1 If you drop your cell phone from the top of your desk as you try to hide it (0.82 m), how long will it take to hit the ground?

Reviewing the Math #2 Find the two components of the velocity for each vector: (vx and vy) v = 25 m/s horizontally   v = 25 m/s at 20° N of E

Motion of Objects Projected Horizontally Chapter 3-3 (Part 1) Motion of Objects Projected Horizontally

Objective 1 Understand how to separate a motion vector into its horizontal and vertical components Objective 2 Understanding that motions which are perpendicular to each other are also independent of each other Objective 3 Solve problems involving objects projected horizontally

Projectiles Projectile: Projectile Motion: objects thrown or launched into the air and subject to gravity Projectile Motion: motion through the air without a propulsion

Projectiles Curved Motion: Resolution into Components: path is parabolic (larger curve for higher launches) Resolution into Components: projectile motion can be separated into both vertical and horizontal components Initial vertical component = viy = 0m/s (object in free-fall, a = g = -9.8m/s2) Initial horizontal component = vix = vx = shot at…

Components of a Horizontal Projectile vx (horizontal velocity) is constant and there is no acceleration in this direction (we ignore air resistance) vy is not constant, it is accelerated downward due to gravity (9.8 m/s2 down or -9.8 m/s2, where the - sign means down)

This means: In the horizontal direction, equal distances are covered in equal amounts of time In the vertical direction, there is acceleration, so different distances are covered in equal amounts of time

y v0 x

y x

y x

y x

y x

g = -9.81m/s2 y Object is in free-fall Acceleration is constant, and downward ay= g = -9.81m/s2 The horizontal (x) component of velocity is constant The horizontal and vertical motions are independent of each other, but they have a common time g = -9.81m/s2 x

WHAT DOES THIS MEAN? Remember: vy = aΔt What are the components at each second? t = 0s 1s 2s 3s 4s

Horizontal and vertical velocities are completely independent of one another!

ax = 0 ay = g = -9.81 m/s2 vi = vx vy = ayΔt vyf2 = 2ayΔy Δx = vxΔt Equations of motion: horizontal (x) uniform motion vertical (y) accelerated motion acceleration ax = 0 ay = g = -9.81 m/s2 velocity vi = vx vy = ayΔt vyf2 = 2ayΔy displacement Δx = vxΔt Δy = ½ ayΔt2 Frame of reference: x y vi h or Δy g or a

Analysis of Motion ASSUMPTIONS: x-direction (horizontal): uniform motion y-direction (vertical): accelerated motion no air resistance POSSIBLE QUESTIONS: What is the total time of the motion? What is the horizontal range (Δx)? What is the initial velocity? What is the final velocity?

How long does it take the football to reach the bottom of the cliff? A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high How long does it take the football to reach the bottom of the cliff? vx= 5m/s 78.4 m

How long does it take the football to reach the bottom of the cliff? A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high How long does it take the football to reach the bottom of the cliff? Unknown: Δt Given: yf = 78.4m yi = 0 m vyi = 0 m/s vxi = 5.0 m/s a = -9.8m/s2 vx= 5m/s Equation: Δy = ½ aΔt2 Solving for Δt Δt2 = 2 Δyf/a Δt = √ 2Δyf/a Δt = √ 2(-78.4m)/(-9.8m/s2) Δt = 4.0 s 78.4 m

How far from the base of the cliff does the football land? A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high How far from the base of the cliff does the football land? vx= 5m/s 78.4m ?

How far from the base of the cliff does the football land? A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high How far from the base of the cliff does the football land? Unknown: Δx Given: yf = 78.4m yi = 0 m vyi = 0 m/s vxi = 5.0 m/s a = -9.8 m/s2 Δt = 4.0 s vx= 5m/s Equation: Δ x = vx Δt 78.4m Solving for Δ x Δ x = (5.0m/s)(4.00s) Δ x = 20. m ?

A football is thrown horizontally at a speed of 5 A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high What are the horizontal and vertical components of the velocity just before it hits the ground (4sec)? vx= 5m/s 78.4m vf

A football is thrown horizontally at a speed of 5 A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high What are the horizontal and vertical components of the velocity just before it hits the ground (4sec)? Unknown: vfx , vfy Given: vxi =5.0 m/s yi = 0 m yf = 78.4m vyi = 0 m/s Δt = 4.0 s a = -9.8 m/s2 vx= 5m/s Equation for vxf vxf = vxi Solving vxf = 5.0 m/s Equation for vyf vyf = a Δt Solving vyf = (-9.8 m/s2 )(4s) vyf = -39 m/s 78.4m vf

Sample Problem Someone is being chased down a river by someone else in a faster craft. Just as the fast boat pulls up to the slower boat, both reach the edge of a 5.0 m waterfall. If the slower boat’s velocity is 15 m/s and the faster boat’s speed is 26 m/s, how far apart will the two vessels be when they land?

sketch known(x) known(y) unknown equation/solution known(x) known(y) unknown equation/solution

Sample Problem An African Spitting Cobra can raise its head straight up approximately 0.61m. An average distance that the poisonous spit travels is 3.60m. What is the horizontal velocity of this deadly venom? sketch known(x) known(y) unknown equation/solution

Sample Problem A girl jumps off of the 10. m platform with a horizontal velocity of 2.0 m/s. How far from the end of the platform does she hit the water?

(Δt = 1.43 sec)

Sample Problem An army helicopter needs to drop supplies to troops in the field. If the army helicopter is flying at an altitude of 500 m and a horizontal velocity of 10 m/s, how far before it gets to the target zone should they drop the supplies?

(Δx = 101m)