When we throw a ball :

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1 PROJECTILE MOTION

2 When we throw a ball : There is a constant velocity horizontal motion And there is an accelerated vertical motion These components act independently of each other

3 PROJECTILE MOTION A falling object with constant linear velocity and vertical acceleration :

4 2-D motion The path or trajectory projectiles make is parabolic (neglecting air resistance). Two independent motions- horizontal and vertical. Use kinematics equations in one direction at a time. The connection between the two motions is the variable time.

5 Projectile motion Vertical motion is free fallconstant acceleration motion. becomes Dx = v i t + ( ½) a t 2 v f = v i + at v f2 = v i2 +2aDx

6 Projectile motion Horizontal motion is constant velocity motion. v = v = x f xi v x..and a x = 0, so Dx = v i t + ( ½) a t 2 becomes Dx = v ix t

7 A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile. Problem Type 1:

8 Problem Type 2: A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion which is symmetrical to its path upwards to the peak. Predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it is at its peak.

9 Example Problem 1 A projectile is launched from a height of 44.1 m with a initial horizontal speed of 20 m/s. a) How long is it the air? b) how far does it travel horizontally before it hits the ground? Horizontal: v xi = 20 m/s Vertical: v yi = 0 m/s Dy = m a y = m/s 2 a) Dy = v yi Dt + ½ a y Dt m = (0 m/s) Dt + 1/2 (-9.8 m/s 2 ) Dt 2 Dt = Ö (2 (-44.1 m) / -9.8 m/s 2 ) = 3 s m 20m/s b) Dx = v x Dt Dx = (20 m/s) (3s) = 60m

10 Example 1 Continued c) At what velocity does the projectile hit the ground? t= 3 s; Horizontal: v xi = 20 m/s Vertical: v yi = 0 m/s Dy = m a y = m/s v yf = v yi + a y Dt v yf = (0 m/s) + (-9.8 m/s 2 ) (3s) = m/s v yf q v x v v = Ö (v yf2 + v x2 ) = Ö( m/s) 2 + (20m/s) 2 = 35.6 m/s q = tan -1 (v yf / v x ) below the horizontal q = tan -1 (29.4 m/s / 20 m/s) = below the horizontal v = below the horizontal (or with the horizontal)

11 Problem Type 2

12 Projectiles launched at an angle initial velocity = v i launch angle q You must separate initial velocity into components v yi v i q v x

13 Components from trig. func. Use sine and cosine functions to find components. v yi = v i sin q v xi = v i cos q

14 Example 2: A football is punted with a velocity of 27 m/s at an angle of 30. Find the ball s hang time, maximum height, and distance traveled (range) when it hits the ground. (Assume the ball is kicked from ground level.) Looking for: Total time (t) Max height (y) Range (x) Given: v i = (27m/s, 30 o )

15 Projectile Motion at an angle

16 What do we do with the given info? v i = (27m/s, 30 o ) v i = (23.4m/s, 13.5m/s) What are the units? m/s resolved vector 27m/s V iy = 27sin30 30 o V ix = 27cos30 V iy = 13.5 V iy = 13.5m/s V ix = 23.4 V ix = 23.4m/s

17 So where does this info fit in the chart? Horizontal Vertical a x = v ix = v fx = t = x = m/s 23.4m/s a y = v iy = 13.5m/s v fy = = -13.5m/s t = y =

18 To find max height, deal with only half of the full projectile motion (the upward motion). Up: Horizontal Vertical a x = v ix = v fx = t = x = m/s 23.4m/s a y = - 9.8m/s 2 v iy = 13.5m/s v fy = 0 t = y = v f2 = v i2 + 2gy y = 9.3m For time to reach peak, v f = v i + gt t = 1.38s

19 Projectile Motion at an angle

20 On the horizontal side l Max height occurs midway through the flight. l We found t = 1.38s to get to this max height. l How long is the projectile in the air? DOUBLE this time for total air time t = 1.38(2) = 2.76s l What about range? x = v x t x = (23.4)(2.76) x = 64.6m = RANGE

21 Maximum Range vs. Maximum Height l What angle of a launched projectile gets the maximum height? 90 o l What angle of a launched projectile gets the maximum range? 45 o

22 Projectile Motion at Various Initial Angles l l Complementary values of the initial angle result in the same range The heights will be different The maximum range occurs at a projection angle of 45 o

23 Non-Symmetrical Projectile Motion l l Follow the general rules for projectile motion Break the y-direction into parts up and down symmetrical back to initial height and then the rest of the height

24 Example of Projectile that Launches/Lands at Different Heights

25 Example 3 A stone thrown off a bridge 20 m above a river has an initial velocity of 12 m/s at an angle of 45 degrees above the horizontal. The stone lands in the river below. (a) what is the range of the stone? (b) at what velocity does the stone strike the water?

26 A stone thrown off a bridge 20.0 m above a river has an initial velocity of 12 m/s at an angle of 45 degrees above the horizontal. The stone lands in the river below. (a) what is the range of the stone? (b) at what velocity does the stone strike the water? v iy = 12(sin 45 o ) = 8.49 m/s v ix = 12(cos 45 o ) = 8.49 m/s Dy = m Range means Dx Dx = v ix t In order to avoid solving a quadratic equation for t, let s find v fy v fy 2 =v iy 2 + 2aDy v fy =-21.5 m/s Now find t v fy =v iy + at t = 3.06 sec

27 For range, Dx = v ix t Dx =26 m b) at what velocity does the stone strike the water? = 23.2 m/s v fx v f v fy q below horizontal

28 In the figure below, a stone is projected at a cliff of height h with an initial speed of 42.0 m/s directed at angle θ = 60.0 degrees above the horizontal. The stone strikes at A, 5.50 s after launching. Find (a) the height of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground.

29 In the figure below, a stone is projected at a cliff of height h with an initial speed of 42.0 m/s directed at angle θ = 60.0 degrees above the horizontal. The stone strikes at A, 5.50 s after launching. Find (a) the height of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground. Height of cliff, h= 51.8 m v f = deg below horizontal Maximum height, H= 67.5 m

Chapter 4. Motion in Two Dimensions

Chapter 4. Motion in Two Dimensions Chapter 4 Motion in Two Dimensions Projectile Motion An object may move in both the x and y directions simultaneously. This form of two-dimensional motion we will deal with is called projectile motion.