2. Motion in Two Dimensions Chapter 7.2

3. Projectile Motion What is the path of a projectile as it moves through the air? ? . What forces act on projectiles? Only, which acts only in the y-direction. is ignored in projectile motion.

4. Choosing Coordinates & Strategy For projectile motion: Choose the for motion where is a factor. Choose the for motion. Since there are forces acting in this direction (of course we will neglect friction due to air resistance), the speed will be (a = __). Analyze motion along the y-axis from the x-axis. If you solve for in one direction, you automatically solve for in the other direction.

5. The Trajectory of a Projectile What does the free-body diagram look like for force?

6. The Vectors of Projectile Motion What vectors exist in projectile motion? in both the ___ and ___ directions. in the ___ direction only. v y () v x () a y = a x = Why is the velocity constant in the x-direction?. Why does the velocity increase in the y-direction?

7. Ex. 1: Launching a Projectile Horizontally A cannonball is shot horizontally off a cliff with an initial velocity of 30 m/s. If the height of the cliff is 50 m: How far from the base of the cliff does the cannonball hit the ground? With what speed does the cannonball hit the ground?

8. Diagram the problem ___m F g = F net a = ___ v i = v f = ? vxvx vyvy x = ?

9. State the Known & Unknown Known: x i = 0 v ix = ___ m/s y i = ___ m/s v iy = ___ m/s a = ___ = ___ m/s 2 y = ____ m Unknown: x at y = -50 m v f = ?

10. Perform Calculations (y) Strategy: Use reference table to find formulas you can use.  v fy =  y =  Note that __ has been substituted for __ and __ for __. Use known factors such as in this case where the initial velocity in the y-direction is known to be to simplify the formulas.  v fy =  y =

11. Perform Calculations (y) Now that we have, we can use the first formula to find the final velocity. v fy = v y =

12. Perform Calculations (x) Strategy: Since you know the for the ( ), you also have it for the. is the only variable that can transition between motion in both the __ and __ directions. Since we air resistance and gravity does not act in the (x-direction), a = __. Choose a formula from your reference table x =  Since a = ___, the formula reduces to x = _____ x =

13. Finding the Final Velocity (v f ) We were given the initial x-component of velocity, and we calculated the y-component at the moment of impact. Logic: Since there is no acceleration in the horizontal direction, then ____ = ____ We will use the. v fx = v fy = v f = ?

14. Ex. 2: Projectile Motion above the Horizontal A ball is thrown from the top of the Science Wing with a velocity of 15 m/s at an angle of 50 degrees above the horizontal. What are the x and y components of the initial velocity? What is the ball’s maximum height? If the height of the Science wing is 12 m, where will the ball land?

15. Diagram the problem y x v i = ____ m/s  = ___ ° a y = Ground ___ m x = ? v i = _____ v ix v iy  = ____ °

16. State the Known & Unknown Known: x i = 0 y i = ___ m v i = ___ m/s  = ___° a = ___ = ____ Unknown: y max = ? t = ? x = ? v iy = ? v ix = ?

17. Perform the Calculations (y max ) y-direction: Initial velocity: v iy =  v iy = Time when v fy = 0 m/s: v fy = v iy – gt (ball at )  t = Determine the maximum height: y max =  y max = v i = 15 m/s v xi v yi  = 50 °

18. Perform the Calculations (t) Since the ball will accelerate due to gravity over the distance it is falling back to the ground, the time for this segment can be determined as follows Time from peak to when ball hits the ground:  From reference table: y max =  Since ___ can be set to zero as can ___,  t = By adding the it takes the ball to reach its maximum height ()to the it takes to reach the ground will give you the total.  t total =

19. Perform the Calculations (x) x-direction: Initial velocity:  v ix = Determine the total distance:  x = v i = 15 m/s v xi v yi  = 50°

20. Analyzing Motion in the x and y directions independently. x-direction: d x = v ix t = v fx t d x = v ix t = v fx t v ix = v i  cos  v ix = v i  cos y-direction: d y = ½ (v i + v f ) t = v avg t d y = ½ (v i + v f ) t = v avg t v f = v iy + gt v f = v iy + gt d y = v iy t + ½ g(t) 2 d y = v iy t + ½ g(t) 2 d v fy 2 = v iy 2 + 2gd v iy = v i  sin  v iy = v i  sin 

21. Key Ideas Projectile Motion: is the only force acting on a projectile. Choose a coordinate axis that where the x- direction is along the and the y-direction is. Solve the x and y components. If is found for one dimension, it is also known for the other dimension.