Warm UpWarm Up

1. g = _________1. g = _________ Don’t forget units*Don’t forget units*

2. True or False: In a Vacuum Heavy objects 2. True or False: In a Vacuum Heavy objects fall faster than less massive ones.fall faster than less massive ones.

3. A Badger has a constant acceleration of 6 3. A Badger has a constant acceleration of 6 m/sm/s22 and an initial velocity of 3 m/s, how far and an initial velocity of 3 m/s, how far will the badger go in 25 seconds?will the badger go in 25 seconds?

Warm UpWarm Up

1. What do you think a Frame of reference is?1. What do you think a Frame of reference is? 2. What is a Projectile?2. What is a Projectile? 3. The back wheels of a car mysteriously fall 3. The back wheels of a car mysteriously fall

off just as the car reaches 80 km/h, the car off just as the car reaches 80 km/h, the car comes screeching to a stop 6 s later. What was comes screeching to a stop 6 s later. What was the car’s acceleration?the car’s acceleration?

Reference FrameReference Frame

Physical object which is the basis of our Physical object which is the basis of our coordinates.coordinates. What we are basing our measurements off of. What we are basing our measurements off of.

Projectile MotionProjectile Motion

http://www.mansfieldct.org/schools/mms/sthttp://www.mansfieldct.org/schools/mms/staff/hand/Projectilemotion.htmaff/hand/Projectilemotion.htm

Horizontal and Vertical motions are Horizontal and Vertical motions are independent.independent.

Projectile Motion problemsProjectile Motion problems Step 1: Identify VariablesStep 1: Identify Variables

Known and unknownKnown and unknown Step 2: Set up equationsStep 2: Set up equations

Vertical motion: Vertical motion: a = ga = g VVoyoy = V sin = V sin θθ

Horizontal Motion Horizontal Motion a = 0a = 0 VV0x0x = V cos = V cos θ = Vθ = Vxx

Step 3: Solve equations for unknownsStep 3: Solve equations for unknowns

Projectile Motion VariablesProjectile Motion Variables

V = VelocityV = Velocity VV00 = Initial velocity = Initial velocity VV0x0x = Initial X Velocity = Initial X Velocity VV0y0y = Initial Y velocity = Initial Y velocity VVxx = X Velocity = X Velocity VVyy = Y velocity = Y velocity Y = Height/Vertical Y = Height/Vertical

positionposition

θ = Launch Angleθ = Launch Angle t = timet = time g = acceleration of g = acceleration of

gravity 9.8 m/sgravity 9.8 m/s22 R = RangeR = Range X = Horizontal/X X = Horizontal/X

positionposition

RememberRemember

Vertical and horizontal motions are linked by Vertical and horizontal motions are linked by the variable time.the variable time.

At the initial height: y = 0 (at the beginning At the initial height: y = 0 (at the beginning and end)and end)

At the top of the trajectory: At the top of the trajectory: VVyy = 0 = 0 t is half what it will be when the projectile returns t is half what it will be when the projectile returns

to its initial height.to its initial height.

Projectile motion equationsProjectile motion equations

X-XX-X00 = = VV0x 0x * t* t X-XX-X00 = (V = (V

00 cos cos θ) * tθ) * t Y-YY-Y00 = V = V0Y 0Y ** t – (1/2)*g*tt – (1/2)*g*t22

Y-YY-Y00 = (V = (V00 sin θ) sin θ) ** t – (1/2)*g*tt – (1/2)*g*t22

VVyy = V = V00 sin θ – g*t sin θ – g*t (V(Vyy))22 = (V = (V00 sin θ) sin θ)22 – 2 g (Y-Y – 2 g (Y-Y00))

Path equationPath equation

Y = (tan Y = (tan θ)x – (g xθ)x – (g x22) / ( 2 (V) / ( 2 (V00 cos θ) cos θ)2 2 ))

Note: this is a parabolic pathNote: this is a parabolic path

RangeRange

R = (2 (VR = (2 (V00))22 / g) * sin (2* / g) * sin (2*θ)θ)

Max range is at θ = 45 degreesMax range is at θ = 45 degrees

#41 You take a running leap #41 You take a running leap off a high dive platform. off a high dive platform.

You were running at 2.8 m/s You were running at 2.8 m/s and hit the water 2.6 s later. and hit the water 2.6 s later.

A.A. How high was the platform?How high was the platform?

B.B. How far from the edge of the How far from the edge of the platform did you hit the water?platform did you hit the water?

A quarterback passes the ball with A quarterback passes the ball with an initial velocity of 30 m/s and an initial velocity of 30 m/s and

releases the ball at a 30 degree angle. releases the ball at a 30 degree angle.

A. Sketch the scenario.A. Sketch the scenario. B. What was the maximum height achieved by B. What was the maximum height achieved by

the ball?the ball? Assume the ball is caught by the receiver at Assume the ball is caught by the receiver at

the same height it was thrown fromthe same height it was thrown from C. How long was the ball in the air?C. How long was the ball in the air? D. How long of a completion was it?D. How long of a completion was it?

43. A pitched ball is hit by a 43. A pitched ball is hit by a batter at a 45 degree angle and batter at a 45 degree angle and just clears the outfield fence, 98 just clears the outfield fence, 98 m away. If the fence was at the m away. If the fence was at the same height as the ball when it same height as the ball when it was hit, what was the velocity was hit, what was the velocity of the ball when it left the bat?of the ball when it left the bat?

HomeworkHomework

Pg 172 (7, 19, 22, Pg 172 (7, 19, 22, 23, 39, 40, 42, 45)23, 39, 40, 42, 45)

Have a great day!Have a great day!