1. Angled Projectiles

2. Projectiles launched @ an angle These projectiles are different from those launched horizontally since they now have an initial vertical velocity Examples: Kicking a football or a soccer ball Throwing a baseball Hitting a golf ball off a tee Launching a cannonball from a cannon

3. Components of Initial Velocity Horizontal Component v ix = v i cos θ Vertical Component v iy = v i sin θ V i = Initial Velocity Θ = angle the projectile is launched Remember: SOH CAH TOA Imagine the horizontal and vertical components are on an x and y axis, respectively. In this case, the resultant of the two vectors would be the initial velocity and it would also act as the “hypotenuse” of the triangle that the vector resolution creates.

4. Comparison of v x and v y Horizontal velocity (v x ) stays constant during path Initial vertical velocity (v iy ) is positive Vertical velocity (v y ) at maximum height is zero Final vertical velocity (v fy ) is negative

5. Maximum Range vs. Maximum Height What angle of a launched projectile gets the maximum height? What angle of a launched projectile gets the maximum range? These can be determined by comparing the initial horizontal and vertical velocities. 90 o 45 o

6. Longest Range A 45˚ will provide the longest horizontal range.

7. Highest height A 90 ˚ angle will produce the highest maximum height

8. Sample Question #1 (finding initial velocities both x and y) A cannonball is fired from ground level at an angle of 60° with the ground at a speed of 72 m/s. What are the vertical and horizontal components of the velocity at the time of launch?

9. Sample Question #1 Answer and Explanation Given v i = 72 m/s θ = 60° Calculations: v ix = v i cos θ v ix = 72 m/s cos60° v ix = 36 m/s v iy = v i sin θ v iy = 72 m/s sin60° v iy = 62.35 m/s 60˚ v iy = v i sinΘ v ix = v i cosΘ

10. Sample Question #2 (finding maximum height) You kick a soccer ball at an angle of 40° above the ground with a velocity of 20 m/s. What is the maximum height the soccer ball will reach? 40˚ 20 m/s

11. Sample Question #2 Answer and Explanation Given v i = 20 m/s θ = 40° For this question, you must find the initial vertical velocity in order to find the maximum height. Then plug it into a second equation noting that v fy = 0 m/s at maximum point. v iy = v i sin θ v iy = 20m/s sin40° v iy = 12.86 m/s v fy 2 = v iy 2 – 2a Δy (0m/s) 2 = (12.86 m/s) 2 – 2(9.8 m/s 2 ) Δy -165.38 m 2 /s 2 = (-19.6m/s 2 ) Δy 8.44 m = Δy

12. Sample Question #3 A cannonball is launched at ground level at an angle of 30° above the horizontal with an initial velocity of 26 m/s. How far does the cannonball travel horizontally before it reaches the ground? 30˚ 26 m/s

13. Sample Question #3 Answer and Explanation v ix = v i cos θ v ix = 26 m/s cos30° v ix = 22.56 m/s v iy = v i sin θ v iy = 26 m/s sin30° v iy = 13 m/s For this question, the vertical velocity must be found to find the total time. The horizontal velocity will need to be found to find the total range. X MotionY Motion xixi 0 m xfxf vivi vfvf a 0 m/s 2 -9.8 m/s 2 t 22.56 m/s 13 m/s -13 m/s 2.653 s Step 2: Find time a= v fy - v iy t -9.8 m/s 2 = -13 m/s – 13 m/s t t = 2.653 s Step 3: Find range v x = Δx t 22.56 = Δ x 2.653 Δx = 59.85 m

14. Sample Question #4 The punter for the Panthers punts the football with a velocity of 17 m/s at an angle of 25 . Find the ball’s hang time, distance traveled (range), and maximum height when it hits the ground. (Assume the ball is kicked from ground level.) 25˚ 17 m/s

15. Step 1: Find the velocity components v ix = v i cos θ v ix = 25 m/s cos55° v ix = 14.34 m/s v iy = v i sin θ v iy = 25 m/s sin55° v iy = 20.48 m/s Sample Question #4 The punter for the Panthers punts the football with a velocity of 25 m/s at an angle of 55 . Find the ball’s hang time, distance traveled (range), and maximum height, when it hits the ground. (Assume the ball is kicked from ground level.) X MotionY Motion xixi 0 m xfxf vivi vfvf a 0 m/s 2 -9.8 m/s 2 t 14.34 m/s 20.48 m/s -20.48 m/s 4.179 s 59.93 m Step 2: Find time a= v fy - v iy t -9.8 m/s 2 = -20.48 m/s – 20.48 m/s t t = 4.179 s Step 3: Find range v x = Δ x t 14.34 m/s = Δ x 4.179 s Δ x = 59.93 m Step 4: Find maximum height Δy = v iy t + ½at 2 Δy = 20.48(2.090) + ½ (-9.8)(2.090 2 ) Δy = 42.79 + (-21.40) Δy = 21.40 m Or v fy 2 = v iy 2 + 2a Δy 0 = 20.48 2 + 2(-9.8)(Δy) 0 = 419.4 + -19.6 Δy -419.4 = -19.6 Δy Δy = 21.40 m

16. your homework … We are going to see what kind of job Hollywood writers and producers would do on their Watching a clip of the Bus Jump, use the timer provided to time the flight of the bus and then do the actual calculations for homework.Bus Jump