2Objectives: 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.
3Projectile MotionA projectile is a particle moving near the Earth’s surface under the influence of its weight only (directed downward).WWWa = g
4Vertical and Horizontal Motion Simultaneously dropping yellow ball and projecting red ball horizontally.Click right to observe motion of each ball.
5Vertical 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.
6Ball Projected Horizontally and Another Dropped at Same Time: voxVertical Motion is the Same for Each Ball1 s2 s3 svyvx
7Observe Motion of Each Ball voxVertical Motion is the Same for Each Ball1 s2 s3 s
8Consider 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.
9Displacement Calculations for Horizontal Projection: For any constant acceleration:For the special case of horizontal projection:Horizontal displacement:Vertical displacement:
10Velocity Calculations for Horizontal Projection (cont.): For any constant acceleration:For the special case of a projectile:Horizontal velocity:Vertical velocity:
11First 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
12Example 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
13Consider 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
14Displacement Calculations For General Projection: The components of displacement at time t are:For projectiles:Thus, the displacement components x and y for projectiles are:
15Velocity Calculations For General Projection: The components of velocity at time t are:For projectiles:Thus, the velocity components vx and vy for projectiles are:
16Problem-Solving Strategy: Resolve initial velocity vo into components:vovoxvoyq2. Find components of final position and velocity:Displacement:Velocity:
17Problem 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.
18Example 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
19Example 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
20Example 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 (+)
22Vertical 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:
23Example 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
24Example 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 =
29Example 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
30Example 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
31The 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)
32Example 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);
33Example 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
34The 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
35Example 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 =
36Vox = (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.
37Example 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
38Draw 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
39Range 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
40Summary for Projectiles: 1. Determine x and y components v02. The horizontal and vertical components of displacement at any time t are given by:
41Summary (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: