 # Chapter 5 Projectile motion. 1. Recall: a projectile is an object only acted upon by gravity.

## Presentation on theme: "Chapter 5 Projectile motion. 1. Recall: a projectile is an object only acted upon by gravity."— Presentation transcript:

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

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

5. We will separate curved motion into independent x and y motions

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

7. Observe: a large ball bearing is dropped at the same time as a second ball bearing is fired horizontally.

What happens?

Remember adding 2 perpendicular vectors horizontal and vertical vectors.

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

Boat in a river

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

Traveling up and down stream

10. How fast will a boat drift downstream?

11. V total = V boat + V current V total = 0 + 8 = 8 m/sec

Now using the motor…

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

12. Suppose the motor moves the boat at 15 m/sec. how fast will the boat travel downstream [with the current]?

13. Using motor with current: V total = V boat + V current

Total velocity traveling downstream: V total = V boat + V current = 15 + 8 = 23 m/sec 

14.Using motor against the current: total velocity traveling upstream [AGAINST the current] Vtotal = Vboat –Vcurrent

15. Going upstream: Velocity of current and boat are opposite directions Vtotal = Vboat –Vcurrent

Vtotal = 15 -8 = 7 m/sec 

Crossing the river

16. If there was no current, how many seconds for the boat to travel 120 meters to reach the opposite side?

Velocity = distance time or time = distance velocity

time = distance velocity time = 120 m 15 m/sec time = 8 seconds

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

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

19. Add the perpendicular velocity vectors add to find the resultant velocity

The triangles are similar:

20. In this example, Every second the boat travels 15 meter in the x direction IT ALSO TRAVELS 8 meter in the y direction

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

Velocity = distance time so distance = velocity X time distance = 8 m/sec X 8 sec = 64 m

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

21) If you want to go from A to B instead, you must point the boat diagonally upstream to compensate for the current.

22) Planes are affected by the wind the same way

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

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

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

Break Vboat into Vx andV y components

Use pythagorean theorem to find Vx.

V x 2 + V y 2 = V boat 2 V x 2 + 8 2 = 15 2 V x 2 =225-64 = 161 V x =12.7 m/sec

How many seconds to cross? Velocity = Distance/time T = D/V

How many seconds to cross? From A to B T = D/V T = 120/12.7 T = 9.4 sec

PROJECTILE MOTION

Projectile motion: A projectile that has horizontal motion has a parabolic trajectory We can separate the trajectory into x motion and y motion.

In the x direction: constant velocity V x = constant distance in x direction X = V x t

In y direction: free fall = constant acceleration. Velocity in y direction : V = V o – g t Distance in y direction Y = Y o + V o t – ½ g t 2

The range of a projectile is the maximum horizontal distance.

Range and maximum height depend on the initial elevation angle.

If you throw a projectile straight up, the range = 0 height is maximum. 0 degrees : the minimum range but the maximum height.

The maximum range occurs at elevation 45 o

And for complementary angles 40 and 50 degrees 30 and 60 degrees 15 and 75 degrees 10 and 80 degrees

The range is identical for complementary angles BUT the larger elevation angle gives a greater maximum height.

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

First calculate the height if there was no gravity. If that case, a projectile would follow a straight line path

the projectile is always a distance 5t 2 below this line.

Y = v oy t – ½ gt 2 Y = v oy t – 5t 2 i

summary

Vectors have magnitude and direction Scalars have only magnitude

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

The perpendicular components of a vector are independent of each other.

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]

The path of a boat crossing a stream is diagonal

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.

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

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

For a projectile fired at an angle, the projectile will be 5t 2 below where it would be if there was no gravity.

Download ppt "Chapter 5 Projectile motion. 1. Recall: a projectile is an object only acted upon by gravity."

Similar presentations