chapter 5 projectile motion
DESCRIPTION
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 PresentationTRANSCRIPT
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Chapter 5 Projectile motion
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1. Recall: a projectile is an object only acted upon by
gravity
![Page 3: Chapter 5 Projectile motion](https://reader035.vdocument.in/reader035/viewer/2022062305/568161a3550346895dd15a37/html5/thumbnails/3.jpg)
2.Chapter 4:
[linear motion]straight line motion
that was ONLY vertical or
ONLY horizontal motion
![Page 4: Chapter 5 Projectile motion](https://reader035.vdocument.in/reader035/viewer/2022062305/568161a3550346895dd15a37/html5/thumbnails/4.jpg)
3. Chapter 5:
looks at motion that followsa diagonal path
or a curved path
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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
![Page 18: Chapter 5 Projectile motion](https://reader035.vdocument.in/reader035/viewer/2022062305/568161a3550346895dd15a37/html5/thumbnails/18.jpg)
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
![Page 22: Chapter 5 Projectile motion](https://reader035.vdocument.in/reader035/viewer/2022062305/568161a3550346895dd15a37/html5/thumbnails/22.jpg)
Total velocity traveling downstream: Vtotal = Vboat + Vcurrent
= 15 + 8 = 23 m/sec
![Page 23: Chapter 5 Projectile motion](https://reader035.vdocument.in/reader035/viewer/2022062305/568161a3550346895dd15a37/html5/thumbnails/23.jpg)
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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