2 Objectives: After completing this module, you should be able to: Describe the motion of a projectile by treating horizontal and vertical components of its position and velocity.Solve for position, velocity, or time when given initial velocity and launch angle.
3 Projectile MotionA projectile is a particle moving near the Earth’s surface under the influence of its weight only (directed downward).WWWa = g
4 Vertical and Horizontal Motion Simultaneously dropping yellow ball and projecting red ball horizontally.Click right to observe motion of each ball.
5 Vertical and Horizontal Motion Simultaneously dropping a yellow ball and projecting a red ball horizontally.W WWhy do they strike the ground at the same time?Once motion has begun, the downward weight is the only force on each ball.
6 Ball Projected Horizontally and Another Dropped at Same Time: voxVertical Motion is the Same for Each Ball1 s2 s3 svyvx
7 Observe Motion of Each Ball voxVertical Motion is the Same for Each Ball1 s2 s3 s
8 Consider Horizontal and Vertical Motion Separately: Compare Displacements and Velocitiesvox1 s2 s3 s0 svx1 svy2 svxvyHorizontal velocity doesn’t change.3 svxvyVertical velocity just like free fall.
9 Displacement Calculations for Horizontal Projection: For any constant acceleration:For the special case of horizontal projection:Horizontal displacement:Vertical displacement:
10 Velocity Calculations for Horizontal Projection (cont.): For any constant acceleration:For the special case of a projectile:Horizontal velocity:Vertical velocity:
11 First find horizontal and vertical displacements: Example 1: A baseball is hit with a horizontal speed of 25 m/s. What is its position and velocity after 2 s?25 m/sxy-19.6 m+50 mFirst find horizontal and vertical displacements:x = 50.0 my = m
12 Example 1 Cont.): What are the velocity components after 2 s? 25 m/svxvyv0x = 25 m/s v0y = 0Find horizontal and vertical velocity after 2 s:vx = 25.0 m/svy = m/s
13 Consider Projectile at an Angle: A red ball is projected at an angle q. At the same time, a yellow ball is thrown vertically upward and a green ball rolls horizontally (no friction).qvx = vox = constantvoyvoxvoNote vertical and horizontal motions of balls
14 Displacement Calculations For General Projection: The components of displacement at time t are:For projectiles:Thus, the displacement components x and y for projectiles are:
15 Velocity Calculations For General Projection: The components of velocity at time t are:For projectiles:Thus, the velocity components vx and vy for projectiles are:
16 Problem-Solving Strategy: Resolve initial velocity vo into components:vovoxvoyq2. Find components of final position and velocity:Displacement:Velocity:
17 Problem Strategy (Cont.): 3. The final position and velocity can be found from the components.Rxyqvovoxvoyq4. Use correct signs - remember g is negative or positive depending on your initial choice.
18 Example 2: A ball has an initial velocity of 160 ft/s at an angle of 30o with horizontal. Find its position and velocity after 2 s and after 4 s.voy160 ft/s30ovoxSince vx is constant, the horizontal displacements after 2 and 4 seconds are:x = 277 ftx = 554 ft
19 Example 2: (Continued) 2 s 4 s voy vox 277 ft 554 ft x2 = 277 ft 160 ft/svox30o277 ft554 ftNote: We know ONLY the horizontal location after 2 and 4 s. We don’t know whether it is on its way up or on its way down.x2 = 277 ftx4 = 554 ft
20 Example 2 (Cont.): Next we find the vertical components of position after 2 s and after 4 s. voy= 80 ft/s160 ft/sq0 s3 s2 s1 s4 sg = -32 ft/s2y2y4The vertical displacement as function of time:Observe consistent units.
21 (Cont.) Signs of y will indicate location of displacement (above + or below – origin). voy= 80 ft/s160 ft/sq0 s3 s2 s1 s4 sg = -32 ft/s2y2y496 ft16 ftVertical position:Each above origin (+)
22 Vertical velocity is same as if vertically projected: (Cont.): Next we find horizontal and vertical components of velocity after 2 and 4 s.voy160 ft/svox30oSince vx is constant, vx = 139 ft/s at all times.Vertical velocity is same as if vertically projected:At any time t:
23 Example 2: (Continued) g = -32 ft/s2 vy= 80.0 ft/s v2 v4 0 s 1 s 2 s q0 s1 s2 s3 s4 sAt any time t:v2y = 16.0 ft/sv4y = ft/s
24 Example 2: (Continued) g = -32 ft/s2 vy= 80.0 ft/s 0 s 3 s 2 s 1 s 4 s q0 s3 s2 s1 s4 sg = -32 ft/s2v2v4Moving Up +16 ft/sMoving down -48 ft/sThe signs of vy indicate whether motion is up (+) or down (-) at any time t.At 2 s: v2x = 139 ft/s; v2y = ft/sAt 4 s: v4x = 139 ft/s; v4y = ft/s
25 (Cont.): The displacement R2,q is found from the x2 and y2 component displacements. y2 = 96 ftqx2= 277 ft0 s2 s4 sR2 = 293 ftq2 = 19.10
26 (Cont.): Similarly, displacement R4,q is found from the x4 and y4 component displacements. y4 = 64 ftx4= 554 ftR4t = 4 sR4 = 558 ftq4 = 6.590
27 (Cont.): Now we find the velocity after 2 s from the components vx and vy. voy= 80.0 ft/s160 ft/sq0 s2 sg = -32 ft/s2v2Moving Up +16 ft/sv2x = 139 ft/sv2y = ft/sv2 = 140 ft/sq2 = 6.560
28 (Cont.) Next, we find the velocity after 4 s from the components v4x and v4y. voy= 80.0 ft/s160 ft/sq0 s4 sg = -32 ft/s2v4v4x = 139 ft/sv4y = ft/sv4 = 146 ft/sq2 =
29 Example 3: What are maximum height and range of a projectile if vo = 28 m/s at 300? voy28 m/svox30oymaxvy = 0vox = 24.2 m/svoy = + 14 m/sMaximum y-coordinate occurs when vy = 0:ymax occurs when 14 – 9.8t = 0 or t = 1.43 s
30 Example 3(Cont.): What is maximum height of the projectile if v = 28 m/s at 300? voy28 m/svox30oymaxvy = 0vox = 24.2 m/svoy = + 14 m/sMaximum y-coordinate occurs when t = 1.43 s:ymax= 10.0 m
31 The time of flight is found by setting y = 0: Example 3(Cont.): Next, we find the range of the projectile if v = 28 m/s at 300.voy28 m/svox30ovox = 24.2 m/svoy = + 14 m/sRange xrThe range xr is defined as horizontal distance coinciding with the time for vertical return.The time of flight is found by setting y = 0:(continued)
32 Example 3(Cont.): First we find the time of flight tr, then the range xr. voy28 m/svox30ovox = 24.2 m/svoy = + 14 m/sRange xr(Divide by t)xr = 69.2 mxr = voxt = (24.2 m/s)(2.86 s);
33 Example 4: A ball rolls off the top of a table 1 Example 4: A ball rolls off the top of a table 1.2 m high and lands on the floor at a horizontal distance of 2 m. What was the velocity as it left the table?Note: x = voxt = 2 m1.2 m2 mRy = voyt + ½ayt2 = -1.2 mFirst find t from y equation:½(-9.8)t2 = -(1.2)t = s
34 The ball leaves the table with a speed: Example 4 (Cont.): We now use horizontal equation to find vox leaving the table top.1.2 m2 mRNote: x = voxt = 2 my = ½gt2 = -1.2 mUse t = s in x equation:The ball leaves the table with a speed:v = 4.04 m/s
35 Example 4 (Cont.): What will be its speed when it strikes the floor? Note:1.2 m2 mvxvyt = svx = vox = 4.04 m/svy = vy + gtvy = m/svy = 0 + (-9.8 m/s2)(0.495 s)v4 = 146 ft/sq2 =
36 Vox = (25 m/s) cos 600; vox = 12.5 m/s Example 5. Find the “hang time” for the football whose initial velocity is 25 m/s, 600.vo =25 m/s600y = 0; a = -9.8 m/s2Time of flight tvox = vo cos qvoy = vo sin qInitial vo:Vox = (25 m/s) cos 600; vox = 12.5 m/sVoy = (25 m/s) sin 600; vox = 21.7 m/sOnly vertical parameters affect hang time.
37 Example 5 (Cont.) Find the “hang time” for the football whose initial velocity is 25 m/s, 600. vo =25 m/s600y = 0; a = -9.8 m/s2Time of flight tvox = vo cos qvoy = vo sin qInitial vo:4.9 t2 = 21.7 t4.9 t = 21.7t = 4.42 s
38 Draw figure and find components: Example 6. A running dog leaps with initial velocity of 11 m/s at 300. What is the range?Draw figure and find components:voy = 11 sin 300v = 11 m/sq =300vox = 9.53 m/svoy = 5.50 m/svox = 11 cos 300To find range, first find t when y = 0; a = -9.8 m/s24.9 t2 = 5.50 tt = 1.12 s4.9 t = 5.50
39 Range is found from x-component: Example 6 (Cont.) A dog leaps with initial velocity of 11 m/s at 300. What is the range?Range is found from x-component:voy = 10 sin 310v = 10 m/sq =310vx = vox = m/sx = vxt; t = 1.12 svox = 10 cos 310Horizontal velocity is constant: vx = 9.53 m/sx = (9.53 m/s)(1.12 s) = 10.7 mRange: x = 10.7 m
40 Summary for Projectiles: 1. Determine x and y components v02. The horizontal and vertical components of displacement at any time t are given by:
41 Summary (Continued):3. The horizontal and vertical components of velocity at any time t are given by:4. Vector displacement or velocity can then be found from the components if desired: