Projectile Motion

Conceptual Challenge

Which boat:

• Q1. Has the highest speed?

• Q2. Takes the least time to cross the river?

• Q3. Travels the shortest distance?

Conceptual Challenge

Which boat:

• Q1. Has the highest speed?

• Q2. Takes the least time to cross the river?

• Q3. Travels the shortest distance?

C, Greatest Resultant

A, Greatest Velocity in the x-dir

B, Resultant is directly across the path

Announcements

• Lab is due tomorrow - Please email it to me instead of printing it. [Word or PDF file]. Deadline is 10:30AM tomorrow (4th period). Any emails received after 10:30AM are considered late

2-D Kinematics

• Relative Motion

• Projectile Motion

• Angled Projectiles

Which forces are acting in the x and y directions? [Neglect air resistance]

Which forces are acting in the x and y directions? [Neglect air resistance]

• Y-direction: Net force = gravity -> gravitational acceleration (ay = -9.8m/s/s)

• X-direction: No net force -> no acceleration (ax = 0 m/s/s; neglect air resistance)

X and Y have no effect on each other!

The time it takes to fall depends on height, not horizontal

velocity [tennis ball example]

We can agree on the time it takes to go up and down

Step 1: GivensX-direction Y-direction

Vx = 4.5 m/s ∆x = ? ∆y = -7.5m

Step 1: GivensX-direction (Const. vel) Y-direction (Free fall)

Vx = 4.5 m/s ∆x = ?

∆y = -7.5m ay = -9.8 m/s/s

Vy,o = 0 m/s

Step 2: Equations that might help?

X-direction (Const. vel) Y-direction (Free fall)

Vx = 4.5 m/s ∆x = ?

∆y = -7.5m ay = -9.8 m/s/s

Vy,o = 0 m/s

Step 2: Equations that might help?

X-direction (Const. vel) Y-direction (Free fall)

Vx = 4.5 m/s ∆x = ?

Vx = ∆x/t -> ∆x = Vx*t

∆y = -7.5m ay = -9.8 m/s/s

Vy,o = 0 m/s

Step 2: Equations that might help

X-direction (Const. vel) Y-direction (Free fall)

Vx = 4.5 m/s ∆x = ?

Vx = ∆x/t -> ∆x = Vx*t

we need time!

∆y = -7.5m ay = -9.8 m/s/s

Vy,o = 0 m/s

Step 2: Equations that might help

X-direction (Const. vel) Y-direction (Free fall)

Vx = 4.5 m/s ∆x = ?

Vx = ∆x/t -> ∆x = Vx*t

we need time!

∆y = -7.5m ay = -9.8 m/s/s

Vy,o = 0 m/s

time is the same for x and y!

Step 2: Equations that might help

X-direction (Const. vel) Y-direction (Free fall)

Vx = 4.5 m/s ∆x = ?

Vx = ∆x/t -> ∆x = Vx*t

we need time!

∆y = -7.5m ay = -9.8 m/s/s

Vy,o = 0 m/s

Let’s look for time first: t = ?

Step 3: Calculations

∆y = Vo,y*t + (1/2)at2

[Note: Vo,y = 0 m/s]

∆y = Vo,y*t + (1/2)at2

∆y = (1/2)at2

t = √[(2∆y)/a]

t = √[(2(-7.5m)/(-9.8m/s2]

t = 1.24s

∆y = -7.5m ay = -9.8 m/s/s

Vy,o = 0 m/s

Let’s look for time first:

t = ?

Step 2: Equations that might help

X-direction (Const. vel) Y-direction (Free fall)

Vx = 4.5 m/s ∆x = ?

Vx = ∆x/t -> ∆x = Vx*t

we need time!

∆y = -7.5m ay = -9.8 m/s/s

Vy,o = 0 m/s t = 1.24s

Step 2: Equations that might help

X-direction (Const. vel) Y-direction (Free fall)

Vx = 4.5 m/s ∆x = ?

t = 1.24s

Vx = ∆x/t -> ∆x = Vx*t

∆y = -7.5m ay = -9.8 m/s/s

Vy,o = 0 m/s t = 1.24s

Step 2: Equations that might helpX-direction (Const. vel) Y-direction (Free fall)

Vx = 4.5 m/s ∆x = ?

t = 1.24s

-> ∆x = Vx*t ∆x = (4.5m/s)(1.24s)

= 5.6m

Done!

∆y = -7.5m ay = -9.8 m/s/s

Vy,o = 0 m/s t = 1.24s

Recap1. List out everything in a table (X,Y)

2. We wanted ∆x but we needed time

3. time is the same in x+y, so we solved it in the other axis

4. once we had time, we plugged in to original problem

Practice

Assume a horizontal launch.

t = 1.75

∆x = 35m

1. A ball is thrown horizontally with a speed of 25 m/s off a ledge that is 20 meters high. a. What is the time of flight of the projectile?

b. How far away from the bottom of the ledge does the ball land?

1. A ball is thrown horizontally with a speed of 25 m/s off a ledge that is 20 meters high. a. What is the time of flight of the projectile?

2.02 s

b. How far away from the bottom of the ledge does the ball land? 50.5 m

2. A bomb is released from a plane flying level at an altitude of 20000 m. It lands 40000 m horizontally from where it was released. a. How long was the bomb in the air?

b. What was the plane's speed when the bomb was released?

2. A bomb is released from a plane flying level at an altitude of 20000 m. It lands 40000 m horizontally from where it was released. a. How long was the bomb in the air? 63.89 s

b. What was the plane's speed when the bomb was released? 626.1 m/s

3. A stone thrown horizontally at a speed of 25 m/s from the top of a cliff takes 14.2 seconds to reach the ground. a. How high is the cliff?

b. How far away does the stone hit the ground?

3. A stone thrown horizontally at a speed of 25 m/s from the top of a cliff takes 14.2 seconds to reach the ground. a. How high is the cliff? 988 m

b. How far away does the stone hit the ground? 355 m

4. A bridge is 150 m above a river. If a young boy spits horizontally at a speed of 6 m/s. a. How long will it take the spit to hit the

water below?

b. How far from the base of the bridge will it hit?

4. A bridge is 150 m above a river. If a young boy spits horizontally at a speed of 6 m/s. a. How long will it take the spit to hit the

water below? 5.53 s

b. How far from the base of the bridge will it hit? 33.2 m

5. A ball is thrown horizontally with a speed of 12 m/s and lands a distance of 36 m away. a. How long was the ball in the air?

b. How high was it thrown from?

5. A ball is thrown horizontally with a speed of 12 m/s and lands a distance of 36 m away. a. How long was the ball in the air? 3 s

b. How high was it thrown from? 44 m