chapter 5 projectile motion

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Chapter 5 Projectile motion

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Chapter 5 Projectile motion. 1. Recall: a projectile is an object only acted upon by gravity. 2. Chapter 4: [linear motion] straight line motion that was ONLY vertical or ONLY horizontal motion. 3. Chapter 5: looks at motion that follows a diagonal path or a curved path. - PowerPoint PPT Presentation

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Page 1: Chapter 5  Projectile motion

Chapter 5 Projectile motion

Page 2: Chapter 5  Projectile motion

1. Recall: a projectile is an object only acted upon by

gravity

Page 3: Chapter 5  Projectile motion

2.Chapter 4:

[linear motion]straight line motion

that was ONLY vertical or

ONLY horizontal motion

Page 4: Chapter 5  Projectile motion

3. Chapter 5:

looks at motion that followsa diagonal path

or a curved path

Page 5: Chapter 5  Projectile motion

4. When you throw a baseball, it travels in an curved path.

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5. We will separate curved motion into

independent x and y motions

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6. vertical motion is not affected by the horizontal motion.

And the horizontal motion is not affected by the vertical motion.

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7. Observe: a large ball bearing is dropped

at the same time as a second ball bearing is fired horizontally.

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What happens?

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Remember

adding 2 perpendicular vectors

horizontal and

vertical vectors.

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8. Remember: When we add perpendicular vectors we use Pythagorean theorem to find the

resultant.

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Boat in a river

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9. a river is 120 meters wide and has a current of 8 m/sec.

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9. a river is 120 meters wide and has a current of 8 m/sec.

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Traveling up and down stream

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10. How fast will a boat drift downstream?

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11. Vtotal = Vboat + Vcurrent

Vtotal = 0 + 8 = 8 m/sec

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Now using the motor…

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12. Suppose the motor moves the boat at 15 m/sec.

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12. Suppose the motor moves the boat at 15 m/sec.how fast will the boat travel

downstream [with the current]?

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13. Using motor with current: Vtotal = Vboat + Vcurrent

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Total velocity traveling downstream: Vtotal = Vboat + Vcurrent

= 15 + 8 = 23 m/sec

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14.Using motor against the current: total velocity traveling upstream [AGAINST the current]

Vtotal = Vboat –Vcurrent

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15. Going upstream: Velocity of current and boat are

opposite directions

Vtotal = Vboat –Vcurrent

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Vtotal = 15 -8 = 7 m/sec

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Crossing the river

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16. If there was no current, how many seconds for the boat to travel 120

meters to reach the opposite side?

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Velocity = distance time

or time = distance velocity

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time = distance velocity

time = 120 m 15 m/sec

time = 8 seconds

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But there if is a current, what happens when you try to go straight across the river from A to

B?

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18. If there is a current, The boat still crosses in 8 seconds, but it lands

downstream at point C

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19. Add the perpendicular velocity vectors

add to find the resultant velocity

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The triangles are similar:

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20. In this example, Every second the boat travels 15 meter

in the x direction IT ALSO TRAVELS

8 meter in the y direction

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How far down stream is the boat when it reaches the opposite shore?

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Velocity = distancetime

so distance = velocity X time

distance = 8 m/sec X 8 sec = 64 m

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What if you want to travel straight across and land at B, not C?

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21) If you want to go from A to B instead, you must point the boat

diagonally upstream to compensate for the current.

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22) Planes are affected by the wind the same way

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23) Head wind: slows the plane[opposite direction]

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24. Tail wind: speeds the plane up [same direction]

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25. crosswind: blows plane off course [wind perpendicular to direction of plane]

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Break Vboat into Vx andV y components

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Use pythagorean theorem to find Vx.

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Vx2 + Vy

2 = Vboat2

Vx2 + 8

2 = 15 2

Vx2 =225-64 = 161

Vx =12.7 m/sec

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How many seconds to cross?Velocity = Distance/time

T = D/V

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How many seconds to cross? From A to BT = D/VT = 120/12.7

T = 9.4 sec

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PROJECTILE MOTION

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Projectile motion:

A projectile that has horizontal motion has a parabolic trajectory

We can separate the trajectory into x motion and y motion.

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In the x direction:

constant velocity

Vx = constant

distance in x direction X = Vx • t

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In y direction: free fall = constant acceleration.

Velocity in y direction : V = Vo – g t

Distance in y directionY = Yo + Vot – ½ g t2

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The range of a projectile is the maximum horizontal distance.

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Range and maximum height depend on the initial elevation angle.

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If you throw a projectile straight up,

the range = 0 height is maximum.

0 degrees : the minimum range but the maximum height.

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The maximum range occursat elevation 45o

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And for complementary angles

40 and 50 degrees30 and 60 degrees15 and 75 degrees10 and 80 degrees

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The range is identical for complementary angles

BUT the larger elevation angle gives a greater maximum height.

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The height of a projectile at any time along the path can be calculated.

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First calculate the height if there was no gravity.

If that case, a projectile would follow a straight line path

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the projectile is always a distance 5t2 below this line.

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Y = voy t – ½ gt2 Y = voy t – 5t2

i

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summary

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Vectors have magnitude and direction

Scalars have only magnitude

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The resultant of 2 perpendicular vectors

is the diagonal of a rectangle that has the 2 vectors as the sides.

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The perpendicular components of a vector are independent

of each other.

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The motion of a boat in a stream is the sum of a constant velocity of a boat [x dir] and the

constant velocity of the stream [y dir]

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The path of a boat crossing a stream is diagonal

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The horizontal component of a projectile is constant,

like a ball rolling on a surface with zero friction.

Objects in motion remain in motion at constant speed.

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The vertical component of a projectile is same as for an object in free fall.

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The vertical motion of a horizontally fired projectile is the same as free fall.

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For a projectile fired at an angle, the projectile will be 5t2 below where it would be if there was

no gravity.

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