Projectile fired __________________________ with an initial _________________

3. up at an angle q • Projectile fired __________________________ • with an initial _________________ • Assume no _________________. The only force • acting on the projectile is _________ . This means • the acceleration is ____________, ______________ speed vi air resistance gravity downward -9.81 m/s2 -9.81 m/s2 ≠ 0 vtop ______, atop = ___________ The velocity is always __________ the path tangent to gravity vi v max. height q the range

4. To solve the problem, vi must be ____________ into its horizontal (vix) and vertical (viy) _____________________. vi resolved viy = _______ visinq components q vicosq vix =____________ √(vix2 + viy2) Where: vi = _______________ is the initial speed, and q = __________________ is the angle. tan-1(viy/vix) There are _____ simultaneous motions: For ___ motion, use: _____________________ For ___ motion, use: _____________________ 2 x dx, vx and ax dy, vy and ay y

5. vix • The horizontal motion is determined by ___ = • _______ . Because there is _______ horizontal force, • vix __________________  _____________ x-motion. vicosq no uniform remains constant acceleration: 0 ax = ax t velocity: vfx = vix vfx t dx vixt dx = displacement: t

6. viy visinq B. Vertical motion is determined by ___ = _______ . Because of ____________, the y motion is like a ball thrown _______________ with an initial speed ____ . gravity viy straight up -g ay = acceleration: ay = -9.8 m/s2 -g t velocity: viy + ayt vfy = viy vfy = visinq – 9.8t t displacement: dy viyt + (1/2)ayt2 dy = = visinqt – (1/2)9.8t2 t

7. Ex 1: Ms. Rudd is fired out of a cannon at a speed of 75 m/s and at an angle of 370 to the horizontal. vix = vicosq = (75 m/s)cos370 75 m/s viy 60. m/s = viy = visinq = (75 m/s)sin370 370 45. m/s vix = To determine how high up she goes and how long she is in the air, "pretend" she is fired __________ ___ but with an initial speed = _____ = __________ straight viy up 45. m/s viy = Given: 45. m/s 1st Unknown: dy ay = -9.8 m/s2 2nd Unknown: t 0 vfy =

8. How far up? vf2 = vi2 + 2ad 02 = 452 + 2(-9.81)d 103 m = d How long is she in the air? vf = vi + at 0 = 45+ (-9.8)t 4.6 s = t 0 Because we chose vfy = ___ , this t represents the time to _________________ . To get the total time of flight, we must _____________________ . So, the total time t = _______ s. You could get this time directly if you assume vfy = __________ . Then: rise only double this time 9.2 -45 m/s vf = vi + at

9. -45 = 45 + -9.8 t 9.2 s = t To determine her range, you must assume her x motion is ____________ at vi = ____ = _______ . vix 60. m/s uniform Given: vix = 60. m/s Unknown: dx 0 ax = 9.2 s t = total Notice that the ___________ time is used here! d = vit + (1/2)at2 (60.)9.2 s = + 0 = 550 m

10. air resistance With no ____________________, only the force of ___________ acts on the object: gravity The trajectory (path) is a________________. vi gravity parabola opposite Air resistance acts in the direction _____________ to its velocity. This _____________ its max. height ' and range. decreases The trajectory is _______ ________________________ no air resistance longer a parabola vi gravity

11. Ex 2: A graphical example On way up: horizontal motion -________________ uniform straight up vertical motion –ball thrown________________ parabola combined motion -______________ 3 s 2 s ay = -9.8 m/s2 1 s vi 2 s 1 s 3 s

12. coming down: The motion is exactly the same as that of a projectile which is _______________________ : fired horizontally 3 s 4 s 5 s 6 s 6 s 5 s 4 s 3 s

13. Velocity vectors: going up 3 s 2 s 1 s vi viy 2 s vix 1 s 3 s vy vx resultant velocity  found by adding ____ and ____  is _______________ to the parabola  is = ________ (NOT = ____ ) at the max. height. tangent to vix 0

14. Velocities coming down: 3 s 4 s 5 s 6 s 6 s 5 s 4 s 3 s v symmetry Notice the ______________ with going up

15. q The effect of changing ___ on the trajectory. Assume all are fired with ________________ vi. same speed 900 600 450 300 Which q results in longest range? Which results in highest trajectory? In longest time in air? Which is a parabola? 450 900 900 all are

16. y 100 75 50 25 q 30 60 45 75 0 15 90 As q increases, the ___ component of vi increases. Because of this: total time in air ________________ , and maximum height ______________ increases increases Complementary ________________________ angles have the same range. angle 800 800 neither 100 neither 600 600 300 neither 430 470 470 Range as a function of q, assuming range for 450 is 100. Fill in the rest:

17. Open your 3-ring binder to the Worksheet Table of Contents. Record the title of the worksheet: Projectile Fired at an Angle WS

18. viy = visinq q vix = vicosq vi In sum: vi must be resolved into vix and viy. • Vertical (y) motion: • use viy to find: • how long it’s in the air • how high it goes • accelerated motion • y motion is like a ball • straight up • Horizontal (x) motion: • use vix to find how far • it travels horizontally • each second • uniform x motion

19. next few slides are from an old version

20. Up and down together: uniform Horizontal motion is ____________________ at vix. Vertical motion is like a ball tossed straight up at viy.

21. At top, what are v and a? What were v and a for the horiz. fired case at top? How does tup compare to tdown? How does ttotal compare to tup or tdown? How do the vup’s compare to vdown’s? Think: mirror _____________about half way point. Second half is same as _______________________ case. The shape of the trajectory is a ___________. Resultant v is always __________ to parabola. v = vixa = -9.8 m/s2 v = 0 a = -9.8 m/s2 equal ttotal=2tup=2tdown equal symmetry horizontally fired parabola tangent

22. Use the equations of motions to predict the position and velocity of the projectile at later times. vf = vi +at d = vit + ½ at2 • Since the object moves in 2 dimensions, each d, v • and a must be replaced by their components: • For x motion: d, v, a  dx, vx and ax • For y motion: d, v, a  dy, vy and ay This is different from the horizontally fired case • vi is a mixture of horizontal: vix= vicosq • and vertical: viy = visinq 3. After the projectile is launched, the only force acting on it is gravity, downward. There is no horizontal force. Because of this, the only acceleration a is purely vertical: ay = -9.8 m/s2 ax = 0

23. Horizontal (x) motion: ax = 0 displacement: d = vit + ½ at2 dx dx = vixt + ½axt2 dx = vixt + ½(0)t2 dx = vixt t velocity: vf = vi +at vfx vfx = vix + axt vfx = vix + (0)t t vfx = vix The x motion is uniformat a speed = vix = vicosq

24. Vertical (y) motion: viy = 0 &ay = -9.8 m/s2 displacement: d = vit + ½ at2 |dy| dy = viyt + ½ayt2 dy = (0)t + ½(-9.8 m/s2)t2 dy = -4.9t2 t velocity: vf = vi +at |vfy| vfy = viy + ayt vfy = 0 + (-9.8 m/s2)t t vfy = - 9.8t The y motion is same as for a dropped ball.

25. Ex: A ball is fired up at an angle as shown at right. viy = 40 q vix = 60 m/s Use g = 10 m/s2 How much time does it spend going up? 4 s 4 s How much time does it spend coming down? How much total time does it spend in the air? 8 s What is its range? dx = vix x ttotal = 60 m/s x 8 s dy = viy x tup = How high up does it go? 20 m/s x 4 s Find the angle q? q = tan-1(40/60) = 340 What are its v and a at maximum height? 60 m/s & -9.8 m/s2