Projectile MotionFalling things, and rockets ‘n’ that…

Physics Unit 5Lesson 1

Projectile MotionIf an object is launched at an initial angle of θ0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component.Initial Vertical Velocity does NOT = ZERO t = 2 Vy /g (if Y = 0)

Initial Horizontal Velocity does NOT = Zero

It can be understood by analyzing the horizontal and vertical motions separately.

Initial Vertical Velocity = ZEROt = √{2y/g} (if Vy =0)

Initial Horizontal Velocity = Zero

Projectile Motion

Projectile Motion EquationsHorizontal Component:X = Vx(t)Vx = Vi COS

Vertical Component:Y = Vy(t) –(1/2)gt2

Vy = Vi SIN

t = 2 Vy /g (if Y = 0)t = √{2y/g} (if Vy =0)

Projectile Motion 2-D Lesson 1 Homework– Glencoe Page 150-151

#’s 1 - 5 ALL

Be able to gather information / data

solve for X and Y Component Velocities solve for Range {X – displacement}solve for Time of Flightsolve for Height {Y – Displacement}

Objectives:

Do Now:

HOMEWORK LESSON 2 – PAGE 164 – 165#45,52,57,59

How far above the ground will a ball be above the ground, if it is thrown from ground level @ 30.625 m/s straight up in 2.50 seconds? Use 9.81 m/s2 for gravity Y = Vi (t) – ½ gt2

Y = 30.625 (2.5) – ½ (9.81)(2.5)2

Y = 76.5625 – 30.625 Y = 45.9375 m = 45.9 m

:

Unit 5 Lesson 2:

Review - Projectile Motion Equations

Horizontal Component:X = Vx(t)Vx = Vi COS

Vertical Component:Y = Vy(t) –(1/2)gt2

Vy = Vi SIN

t = 2 Vy /g (if Y = 0)t = √{2y/g} (if Vy =0)

Projectile Motion• Projectile motion follows a parabolic path i.e. http://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.html • It has a horizontal (x) component and a

vertical (y) component.• The horizontal component is not affected by

gravity, so use separate equations for the different components.

Continued…• What does this mean for objects dropped from

the same height?• Objects dropped from the same height will hit the

ground at the same time, regardless of weight (ignoring differences in air resistance)

http://www.animations.physics.unsw.edu.au/jw/projectiles.htm#1

• Solve the same way you would any other kinematics problem… AICR

• State what you know, what do you want to find?• Choose an equation and solve for your unknown.• Look out for key words like.. “dropped”, “from rest” and “stationary:”

Solving Problems Involving AICR

Projectile Motion

1. Read the problem carefully, and choose the object(s) you are going to analyze.2. Draw a diagram.3. Choose an origin and a coordinate system.4. Decide on the time interval; this is the same in both directions, and includes

only the time the object is moving with constant acceleration g.5. Examine the x and y motions separately.6. List known and unknown quantities. Remember that vx never changes, and that

vy = 0 at the highest point.

7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

Projectile Motion Lesson 3Objectives:Gain proficiency in collecting Data to solve for 2-D motion variables:

Do NOW:How far down range will a cannonball go if fired at an angle of 30.0 deg an initial velocity of 18.5 m/s? assume no air resistance.

HOMEWORK – PAGE: 164 - 165 #’S 42, 45, 51,56 LAB SET UP

What will hit the ground first a bullet dropped or a bullet fired from the same height?Discuss with your neighbor…http://www.youtube.com/watch?v=D9wQVIEdKh8

Objects Under the Influence of Gravity:

 

Projectile Motion

The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g.

This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.

Projectile Motion Day TwoProblemsProjectile Motion Water slideWater Slide 2Giant House Slide

DRIVING OFF A CLIFF!!How fast must the motorcycle leave the cliff to land at x = 90 m, y = -50 m? vx0 = ?

y is positive upwardy0 = 0 at topvy0 = 0

vx = vx0 = ? vy = -gtx = vx0t y = - (½)gt2

Time to Bottom: t = √2y/(-g) = 3.19 svx0 = (x/t) = 28.2 m/s

A PUNT!FIND HANG TIME:RANGE:MAXIMUM HEIGHT:

vx0= v0cos(θ0) = 16.0 15.97 m/svy0= v0sin(θ0) = 12.036 m/sy = y0+ vy0 t – (½)g t2 t= 2.531 sec (quadratic formula)x = vx0 t x = 40.540.49 mt/2 = 1.27 secYmax = y0+ vy0 t – (½)g t2 = 1+ 12.0 (1.27 ) – 4.9 (1.27 )2

Y max = 8.3926 m

v0 = 20.0 m/s, θ0 = 37.0º

Hang Time:2.53 secondsRange: 40.5 mMax Height: 8.39 m

Test Review Unit 5Motion in One-DimensionDisplacement / Instantaneous Velocity / AccelerationMotion DiagramsFree Falling Objects / 1-Dimensional with constant

Acceleration

Motion in Two-Dimensions Vectors Coordinate System / Properties of VectorsComponents of Vectors / Unit VectorsAdding Subtracting Multiplying Vectors (Graphically and

Analytically)Scalar Products - Dot Products – (Honors: Cross Products)

Motion in Two-Dimensions ProjectileRange / Velocity Components / Maximum Height Time of FlightCircular Motion