 Chapter 6B – Projectile Motion

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Chapter 6B – Projectile Motion
A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

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.

Projectile Motion A projectile is a particle moving near the Earth’s surface under the influence of its weight only (directed downward). W W W a = g

Vertical and Horizontal Motion
Simultaneously dropping yellow ball and projecting red ball horizontally. Click right to observe motion of each ball.

Vertical and Horizontal Motion
Simultaneously dropping a yellow ball and projecting a red ball horizontally. W W Why do they strike the ground at the same time? Once motion has begun, the downward weight is the only force on each ball.

Ball Projected Horizontally and Another Dropped at Same Time:
vox Vertical Motion is the Same for Each Ball 1 s 2 s 3 s vy vx

Observe Motion of Each Ball
vox Vertical Motion is the Same for Each Ball 1 s 2 s 3 s

Consider Horizontal and Vertical Motion Separately:
Compare Displacements and Velocities vox 1 s 2 s 3 s 0 s vx 1 s vy 2 s vx vy Horizontal velocity doesn’t change. 3 s vx vy Vertical velocity just like free fall.

Displacement Calculations for Horizontal Projection:
For any constant acceleration: For the special case of horizontal projection: Horizontal displacement: Vertical displacement:

Velocity Calculations for Horizontal Projection (cont.):
For any constant acceleration: For the special case of a projectile: Horizontal velocity: Vertical velocity:

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/s x y -19.6 m +50 m First find horizontal and vertical displacements: x = 50.0 m y = m

Example 1 Cont.): What are the velocity components after 2 s?
25 m/s vx vy v0x = 25 m/s v0y = 0 Find horizontal and vertical velocity after 2 s: vx = 25.0 m/s vy = m/s

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). q vx = vox = constant voy vox vo Note vertical and horizontal motions of balls

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:

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:

Problem-Solving Strategy:
Resolve initial velocity vo into components: vo vox voy q 2. Find components of final position and velocity: Displacement: Velocity:

Problem Strategy (Cont.):
3. The final position and velocity can be found from the components. R x y q vo vox voy q 4. Use correct signs - remember g is negative or positive depending on your initial choice.

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. voy 160 ft/s 30o vox Since vx is constant, the horizontal displacements after 2 and 4 seconds are: x = 277 ft x = 554 ft

Example 2: (Continued) 2 s 4 s voy vox 277 ft 554 ft x2 = 277 ft
160 ft/s vox 30o 277 ft 554 ft Note: 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 ft x4 = 554 ft

Example 2 (Cont.): Next we find the vertical components of position after 2 s and after 4 s.
voy= 80 ft/s 160 ft/s q 0 s 3 s 2 s 1 s 4 s g = -32 ft/s2 y2 y4 The vertical displacement as function of time: Observe consistent units.

(Cont.) Signs of y will indicate location of displacement (above + or below – origin).
voy= 80 ft/s 160 ft/s q 0 s 3 s 2 s 1 s 4 s g = -32 ft/s2 y2 y4 96 ft 16 ft Vertical position: Each above origin (+)

Vertical velocity is same as if vertically projected:
(Cont.): Next we find horizontal and vertical components of velocity after 2 and 4 s. voy 160 ft/s vox 30o Since vx is constant, vx = 139 ft/s at all times. Vertical velocity is same as if vertically projected: At any time t:

Example 2: (Continued) g = -32 ft/s2 vy= 80.0 ft/s v2 v4 0 s 1 s 2 s
q 0 s 1 s 2 s 3 s 4 s At any time t: v2y = 16.0 ft/s v4y = ft/s

Example 2: (Continued) g = -32 ft/s2 vy= 80.0 ft/s 0 s 3 s 2 s 1 s 4 s
q 0 s 3 s 2 s 1 s 4 s g = -32 ft/s2 v2 v4 Moving Up +16 ft/s Moving down -48 ft/s The signs of vy indicate whether motion is up (+) or down (-) at any time t. At 2 s: v2x = 139 ft/s; v2y = ft/s At 4 s: v4x = 139 ft/s; v4y = ft/s

(Cont.): The displacement R2,q is found from the x2 and y2 component displacements.
y2 = 96 ft q x2= 277 ft 0 s 2 s 4 s R2 = 293 ft q2 = 19.10

(Cont.): Similarly, displacement R4,q is found from the x4 and y4 component displacements.
y4 = 64 ft x4= 554 ft R4 t = 4 s R4 = 558 ft q4 = 6.590

(Cont.): Now we find the velocity after 2 s from the components vx and vy.
voy= 80.0 ft/s 160 ft/s q 0 s 2 s g = -32 ft/s2 v2 Moving Up +16 ft/s v2x = 139 ft/s v2y = ft/s v2 = 140 ft/s q2 = 6.560

(Cont.) Next, we find the velocity after 4 s from the components v4x and v4y.
voy= 80.0 ft/s 160 ft/s q 0 s 4 s g = -32 ft/s2 v4 v4x = 139 ft/s v4y = ft/s v4 = 146 ft/s q2 =

Example 3: What are maximum height and range of a projectile if vo = 28 m/s at 300?
voy 28 m/s vox 30o ymax vy = 0 vox = 24.2 m/s voy = + 14 m/s Maximum y-coordinate occurs when vy = 0: ymax occurs when 14 – 9.8t = 0 or t = 1.43 s

Example 3(Cont.): What is maximum height of the projectile if v = 28 m/s at 300?
voy 28 m/s vox 30o ymax vy = 0 vox = 24.2 m/s voy = + 14 m/s Maximum y-coordinate occurs when t = 1.43 s: ymax= 10.0 m

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. voy 28 m/s vox 30o vox = 24.2 m/s voy = + 14 m/s Range xr The 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)

Example 3(Cont.): First we find the time of flight tr, then the range xr.
voy 28 m/s vox 30o vox = 24.2 m/s voy = + 14 m/s Range xr (Divide by t) xr = 69.2 m xr = voxt = (24.2 m/s)(2.86 s);

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 m 1.2 m 2 m R y = voyt + ½ayt2 = -1.2 m First find t from y equation: ½(-9.8)t2 = -(1.2) t = s

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 m 2 m R Note: x = voxt = 2 m y = ½gt2 = -1.2 m Use t = s in x equation: The ball leaves the table with a speed: v = 4.04 m/s

Example 4 (Cont.): What will be its speed when it strikes the floor?
Note: 1.2 m 2 m vx vy t = s vx = vox = 4.04 m/s vy = vy + gt vy = m/s vy = 0 + (-9.8 m/s2)(0.495 s) v4 = 146 ft/s q2 =

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/s 600 y = 0; a = -9.8 m/s2 Time of flight t vox = vo cos q voy = vo sin q Initial vo: Vox = (25 m/s) cos 600; vox = 12.5 m/s Voy = (25 m/s) sin 600; vox = 21.7 m/s Only vertical parameters affect hang time.

Example 5 (Cont.) Find the “hang time” for the football whose initial velocity is 25 m/s, 600.
vo =25 m/s 600 y = 0; a = -9.8 m/s2 Time of flight t vox = vo cos q voy = vo sin q Initial vo: 4.9 t2 = 21.7 t 4.9 t = 21.7 t = 4.42 s

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 300 v = 11 m/s q =300 vox = 9.53 m/s voy = 5.50 m/s vox = 11 cos 300 To find range, first find t when y = 0; a = -9.8 m/s2 4.9 t2 = 5.50 t t = 1.12 s 4.9 t = 5.50

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 310 v = 10 m/s q =310 vx = vox = m/s x = vxt; t = 1.12 s vox = 10 cos 310 Horizontal velocity is constant: vx = 9.53 m/s x = (9.53 m/s)(1.12 s) = 10.7 m Range: x = 10.7 m

Summary for Projectiles:
1. Determine x and y components v0 2. The horizontal and vertical components of displacement at any time t are given by:

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:

CONCLUSION: Chapter 6B Projectile Motion

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