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 2D 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. Δx = v i t + ( ½) a t 2 becomes v f = v i + at v f 2 = v i 2 + 2aΔx
6 Projectile motion Horizontal motion is constant velocity motion...and a x = 0, so Δx = v i t + ( ½) a t 2 becomes
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 Type 1 examples A pool ball leaves a 0.60meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.
10 A pool ball leaves a 0.60meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location. Horizontal (x) info: List Givens Vertical (y) info: x =? v ix = 2.4 m/s a x = 0 m/s 2 y = m v iy = 0 m/s a y = 9.8 m/s 2 t =? Equations: Δx = v i t + ( ½) a t 2 v f = v i + at v f 2 = v i 2 + 2aΔx
11 Choose an equation and solve y = v iy t + ( ½) a y t m = (0 m/s) t (9.8 m/s/s) t m = (4.9 m/s/s) t s 2 = t 2 t = s
12 Now use equation in horizontal direction to solve for x x = v ix t + 0.5a x t 2 x = (2.4 m/s)( s) (0) ( s) 2 x = (2.4 m/s) ( s) x = 0.84 m
13 Example B, problem type 1 A soccer ball is kicked horizontally off a 22.0meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
14 A soccer ball is kicked horizontally off a 22.0meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball. List Givens Horizontal (x) info: Vertical (y) info: x = 35 m v ix =? a x = 0 m/s 2 y = m v iy = 0 m/s a y = 9.8 m/s 2 Δx = v i t + ( ½) a t 2 v f = v i + at v f 2 = v i 2 + 2aΔx
15 Solve y = v iy t + 0.5a y t 2 t= 2.12 s Now use x = v ix t + 0.5a x t 2 to solve for v ix v ix = 16.5 m/s
16 Previous Examples = Problem Type 1
17 Problem Type 2
18 Projectiles launched at an angle initial velocity = v i launch angle θ separate initial velocity into components v yi v i θ v x
19 Components from trig. func. Use sine and cosine functions to find components.
20 Problem type 2 (example A) A football is kicked with an initial velocity of 25 m/s at an angle of 45degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football.
21 back to the problem A football is kicked with an initial velocity of 25 m/s at an angle of 45 o with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football. HORIZONTAL COMPONENT VERTICAL COMPONENT v ix = v i cosθ v ix = 25 m/s cos 45 o v ix = 17.7 m/s v iy = v i sinθ v iy = 25 m/s sin 45 o v iy = 17.7 m/s
22 A football is kicked with an initial velocity of 25 m/s at an angle of 45 o with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football. List Givens Horizontal (x) info: x =??? v ix = 17.7 m/s v fx = 17.7 m/s a x = 0 m/s 2 Vertical (y) info: y =??? v iy = 17.7 m/s v fy = m/s ** a y = 9.8 m/s 2 Equations: Δx = v i t + ( ½) a t 2 v f = v i + at v f 2 = v i 2 + 2aΔx
23 *symmetry of projectile motion
24 Same concept of symmetry applies to free fall motion
25 Use vertical info to find t v fy = v iy + a y t m/s = 17.7 m/s + (9.8 m/s 2 ) t m/s = (9.8 m/s 2 ) t s = t = 3.61 s
26 Now find horizontal distance x = v ix t + ½ a x t 2 x = (17.7 m/s)( s) + ½ (0 m/s/s)( s) 2 x = (17.7 m/s) ( s) x = 63.8 m
27 Find height of projectile at its peak (in other words, the vertical displacement, y) Note: The height is reached when at a time= ½ t y = v iy t + ½ a y t 2 y = (17.7 m/s)(1.80 s) + ½ (9.8 m/s/s)(1.80 s) 2 y = 31.9 m + (15.9 m) y = 15.9 m
28 Find height of projectile at its peak (in other words, the vertical displacement, y) Note: You could also use v f 2 = v i 2 + 2aΔy as long as you know v fy = 0 at the peak height
29 Type 2: example B A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28 o above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the longjumper.
30 A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28 o above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the longjumper. HORIZONTAL COMPONENT VERTICAL COMPONENT v ix = v i cosθ v ix = 12 m/s cos 28 o v ix = 10.6 m/s v iy = v i sinθ v iy = 12 m/s sin 28 o v iy = 5.6 m/s
31 A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the longjumper. Horizontal (x) info: x =??? v ix = 10.6 m/s v fx = 10.6 m/s a x = 0 m/s 2 Equations: List Givens Vertical (y) info: y peak =??? v iy = 5.6 m/s v fy = 5.6 m/s v ypeak = 0 m/s a y = 9.8 m/s 2 Δx = v i t + ( ½) a t 2 v f = v i + at v f 2 = v i 2 + 2aΔx
32 Use vertical info to find t v fy = v iy + a y t 5.6 m/s = 5.6 m/s + (9.8 m/s 2 ) t m/s = (9.8 m/s 2 ) t s = t = 1.1 s
33 Now find horizontal distance x = v ix t + ½ a x t 2 x = (10.6 m/s)( s) + ½ (0 m/s/s)( s) 2 x = (10.6 m/s) ( s) x = 12.2 m
34 Find height of projectile at its peak (in other words, the vertical displacement, y) Note: The height is reached when at a time= ½ t y = v iy t + ½ a y t 2 y = (5.6 m/s)( s) + ½ (9.8 m/s 2 )( s) 2 y = m + ( m) y = 1.6 m
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