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**Chapter 6B – Projectile Motion**

A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

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**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.

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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

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**Vertical and Horizontal Motion**

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

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**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.

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**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

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**Observe Motion of Each Ball**

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

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**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.

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**Displacement Calculations for Horizontal Projection:**

For any constant acceleration: For the special case of horizontal projection: Horizontal displacement: Vertical displacement:

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**Velocity Calculations for Horizontal Projection (cont.):**

For any constant acceleration: For the special case of a projectile: Horizontal velocity: Vertical velocity:

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**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

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**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

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**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

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**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:

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**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:

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**Problem-Solving Strategy:**

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

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**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.

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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

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**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

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**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.

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**(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 (+)

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**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:

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**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

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**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

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**(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

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**(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

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**(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

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**(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 =

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**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

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**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

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**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)

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**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);

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**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

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**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

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**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 =

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**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.

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**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

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**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

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**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

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**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:

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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:

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**CONCLUSION: Chapter 6B Projectile Motion**

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