# Projectile Motion objectives 1.What is a Projectile?What is a Projectile? 2.Characteristics of a Projectile's TrajectoryCharacteristics of a Projectile's.

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Projectile Motion objectives 1.What is a Projectile?What is a Projectile? 2.Characteristics of a Projectile's TrajectoryCharacteristics of a Projectile's Trajectory 3.Describing Projectiles with Numbers a.Horizontal and Vertical Components of VelocityHorizontal and Vertical Components of Velocity b.Horizontal and Vertical Components of DisplacementHorizontal and Vertical Components of Displacement 4.Initial Velocity ComponentsInitial Velocity Components 5.Horizontally Launched Projectiles - Problem-SolvingHorizontally Launched Projectiles - Problem-Solving 6.Non-Horizontally Launched Projectiles - Problem- SolvingNon-Horizontally Launched Projectiles - Problem- Solving

What is a projectile? An object that is launched into the air with some INITIAL VELOCITY Can be launched at ANY ANGLE In FREEFALL after launch (no outside forces except force of gravity) The path of the projectile is a PARABOLA

Free body diagram of a projectile F g, a

Path of a projectile The path of a projectile is parabolic, or arc. The object moves vertically as well as horizontally.

Projectile Motion and Inertia Since a projectile is in free fall after it is launched, the only force acting on it is gravity, which influence the vertical motion of the projectile, causing a vertical acceleration. The horizontal motion of the projectile is the result of inertia. There is no horizontal force. The combination of inertia and gravity causes the parabolic trajectory that is characteristic of projectiles.

Effect of air resistance Ideal projectile Projectile with air resistance

A projectile is an object upon which the only force is ___________. Gravity acts to influence the ___________ motion of the projectile, thus causing a ______________ acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to _____________ in motion at constant velocity. Due to the absence of horizontal forces, a projectile remains in motion with a ______________horizontal velocity. Horizontal forces are _________ required to keep a projectile moving horizontally. The only force acting upon a projectile is _______________ Air resistance _______________ the time, height and range of a projectile. What is projectile motion? – fill in the blanks.

Two types of projectiles Projectile launched horizontally Projectile launched at an angle

Characteristics of a Projectile's Trajectory There are the two components of the projectile's motion – horizontal and vertical motion. These two perpendicular components of motion are independent of each other, which means one component does not affect the other component.perpendicular components of motion are independent of each other

Horizontally Launched Projectile

Non-Horizontally Launched Projectiles

Horizontal components

Vertical components

Projectile components There are two components in projectile motion: horizontal and vertical. These two components are independent of each other. Since the only force acting on the projectile is gravity, which is in the vertical downward direction, it cause vertical acceleration only. The projectile’s vertical motion is the same as if it is in free fall with a y = -g, the vertical velocity changes by -9.81 m/s every second. There is no force acting on the projectile in horizontal direction. The horizontal acceleration is zero. The horizontal motion is only affected by inertia, which means its horizontal velocity is constant.

Horizontal Motion Vertical Motion Forces (Present? - Yes or No) (If present, what dir'n?) No Yes The force of gravity acts downward Acceleration (Present? - Yes or No) (If present, what dir'n?) No Yes "g" is downward at 9.81 m/s/s Velocity (Constant or Changing?) Constant Changing (by -9.81 m/s each second) Horizontal and vertical components are independent of each other. Change of horizontal speed does not affect vertical motion. Change of vertical speed does not affect horizontal motion.

What we know about projectile motion A projectile is any object upon which the only force is gravity, Projectiles travel with a parabolic trajectory due to the influence of gravity, There are no horizontal forces acting upon projectiles and thus no horizontal acceleration, The horizontal velocity of a projectile is constant (a never changing in value), There is a vertical acceleration caused by gravity; its value is 9.81 m/s/s, down, The vertical velocity of a projectile changes by 9.81 m/s each second, The horizontal motion of a projectile is independent of its vertical motion.

Describing Projectiles with Numbers: Horizontal and Vertical Velocity HORIZONTALLY LAUNCHED PROJECTILE Horizontal and Vertical Velocity Horizontal components: vertical components: Object dropped from rest

Timeaxax ayay vxvx vyvy 0 s0-9.81 m/s 2 +20 m/s,0 1 s0-9.81 m/s 2 +20 m/s-9.8 m/s, 2 s0-9.81 m/s 2 +20 m/s-20 m/s, 3 s0-9.81 m/s 2 +20 m/s-29 m/s, 4 s0-9.81 m/s 2 +20 m/s-39 m/s 5 s0-9.81 m/s 2 +20 m/s-49 m/s, Example: cannon ball is launched horizontally at 20. m/s to the right (+) at time t = 0.0 s. Fill in the blanks.

Describing Projectiles with Numbers: Horizontal and Vertical Velocity PROJECTILE LAUNCHED AT AN ANGLE Horizontal and Vertical Velocity Horizontal components: vertical components: Object thrown up and down. v y decreases while up, increase while down.

Timeaxax ayay vxvx vyvy 0 s0-9.81 m/s 2 +73.1 m/s19.6. m/s 1 s0-9.81 m/s 2 +73.1 m/s9.81m/s, 2 s0-9.81 m/s 2 +73.1 m/s0m/s, 3 s0-9.81 m/s 2 +73.1 m/s-9.81 m/s, 4 s0-9.81 m/s 2 +73.1 m/s-19.6 m/s 5 s0-9.81 m/s 2 +73.1 m/s-29.4 m/s, Example: cannon ball is launched with v ix = 73.1 m/s and v iy = 19.6 m/s upward at time t = 0.0 s. Fill in the blanks.

The symmetrical nature of a projectile launched at an angle v y = 0 at top v fy = -v iy t up = t down t total = 2∙t up

Describing Projectiles With Numbers: (Horizontal and Vertical Displacement) HORIZONTALLY LAUNCHED PROJECTILES Horizontal components: vertical components: Object dropped from rest

Timeaxax ayay vxvx vyvy dxdx dydy 0 s0-9.81 m/s 2 +20 m/s,00 m 1 s0-9.81 m/s 2 +20 m/s-9.8 m/s,+20 m-4.91 m 2 s0-9.81 m/s 2 +20 m/s-20 m/s,+40 m-19.6 m 3 s0-9.81 m/s 2 +20 m/s-29 m/s,+60 m-44.1 m 4 s0-9.81 m/s 2 +20 m/s-39 m/s+80 m-78.4m 5 s0-9.81 m/s 2 +20 m/s-49 m/s,+100 m123 m Example: cannon ball is launched horizontally at 20.0 m/s to the right (+) at time t = 0.0 s. Fill in the blanks. The vertical distance fallen from rest during each consecutive second is increasing (i.e., there is a vertical acceleration). The horizontal distance traveled by the projectile each second is a constant value.

Describing Projectiles With Numbers: (Horizontal and Vertical Displacement) PROJECTILES LAUNCHED AT AN ANGLE Horizontal components: vertical components: Object thrown up then come down

Timeaxax ayay vxvx vyvy dxdx dydy 0 s0-9.81 m/s 2 +73.1 m/s19.6. m/s0 m 1 s0-9.81 m/s 2 +73.1 m/s9.81m/s,+73.2 m4.91 m 2 s0-9.81 m/s 2 +73.1 m/s0m/s,+146 m0. m 3 s0-9.81 m/s 2 +73.1 m/s-9.81 m/s,+220 m-4.91 m 4 s0-9.81 m/s 2 +73.1 m/s-19.6 m/s+293 m-19.6m 5 s0-9.81 m/s 2 +73.1 m/s-29.4 m/s,+366 m-44.1 m Example: cannon ball is launched with v ix = 73.1 m/s and v iy = 19.6 m/s upward at time t = 0.0 s. Fill in the blanks. The symmetrical nature of a projectile's trajectory: the vertical displacement of a projectile t seconds before reaching the peak is the same as the vertical displacement of a projectile t seconds after reaching the peak. The horizontal distance traveled by the projectile each second is a constant value.

Horizontal and vertical components are independent of each other The horizontal and vertical motions of a projectile are independent of each other. The horizontal velocity of a projectile does not affect how far (or how fast) a projectile falls vertically. Only vertical motion parameters (initial vertical velocity, final vertical velocity, vertical acceleration) determine the vertical displacement. Only horizontal motion parameters (initial horizontal velocity, final horizontal velocity, horizontal acceleration). Determine the horizontal displacement. One of the initial steps of a projectile motion problem is to determine the components of the initial velocity.

Since velocity is a vector quantity, vector resolution is used to determine the components of velocity. θ vivi v ix v iy SOH CAH TOA sinθ = v iy / v i v iy = v i ∙sinθ cosθ = v ix / v i v ix = v i ∙cosθ Special case: horizontally launched projectile: θ = 0 o v iy = v i sinθ = 0; v ix = v i cosθ = v i Initial Velocity Components

1.A water balloon is launched with a speed of 40 m/s at an angle of 60 degrees to the horizontal. 2. A motorcycle stunt person traveling 70 mi/hr jumps off a ramp at an angle of 35 degrees to the horizontal. 3.A springboard diver jumps with a velocity of 10 m/s at an angle of 80 degrees to the horizontal. Practices – determine horizontal and vertical components of velocity The point of resolving an initial velocity vector into its two components is to use the values of these two components to analyze a projectile's motion and determine such parameters as d x, d y, v fy, t total, t up, etc.

Determination of the TIME OF FLIGHT for projectile launched at an angle, given initial speed and angle: For the given projectile, we can determine the initial horizontal and vertical velocity: v iy = v i sinθ; v ix = v i cosθ For a projectile launched at an angle, its vertical motion is the same as free fall with initial vertical velocity: If we know the initial vertical velocity, we can determine the time to reach the highest the point and the total time of flight. v y = 0 at top; v fy = -v iy t up = t down t total = 2∙t up v fy = v iy + a∙t 0 = v iy - g∙t up Time is the same for both vertical and horizontal components

Determination of HORIZONTAL DISPLACEMENT for projectile launched at an angle, given initial speed and angle and time of flight For the given projectile, we can determine the initial horizontal and vertical velocity: v iy = v i sinθ; v ix = v i cosθ For a projectile launched at an angle, its horizontal motion is constant. To determine its horizontal displacement we can use

Determination of the PEAK HEIGHT for projectile launched at an angle, given initial speed and angle For the given projectile, we can determine the initial horizontal and vertical velocity: v iy = v i ∙sinθ; v ix = v i ∙cosθ For a projectile launched at an angle, its vertical motion is the same as free fall with initial vertical velocity: At the very top, v y = 0 v y 2 = v iy 2 + 2a y ∙y 0 = v iy 2 – 2g∙y peak (a y = -g)

Total flight time, range, max height Time to go up t = v i sinθ / g As θ increases, flight time increase. Max time: θ = 90 o Max height h max = (v i sinθ) 2 /2g As θ increases, flight height increase. Max height: θ = 90 o Range Range = v i 2 sin2θ /g Projectile has maximum range when θ = 45 o

A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football. Horizontal Component Vertical Component v ix = v i cosθ v ix = 25 m/scos45 o v ix = 17.7 m/s v fx = 17.7 m/s a x = 0 m/s/s v iy = v i sinθ v iy = 25 m/ssin45 o v iy = 17.7 m/s v top =0 v fy = -17.7 m/s a y = -9.81 m/s/s Non-Horizontally Launched Projectile Problems t total = 3.61 sx = 63.8 my peak = 15.9 m

example A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper. Horizontal Component Vertical Component v ix = v i cosθ v ix = 12 m/scos28 o v ix = 10.6 m/s v fx = 10.6 m/s a x = 0 m/s/s v iy = v i sinθ v iy = 12 m/ssin28 o v iy = 5.6 m/s v fy = -5.6 m/s a y = -9.8 m/s/s t total = 1.1 sx = 12.2 my peak = 1.6 m

Check your understanding 1.Aaron Agin is resolving velocity vectors into horizontal and vertical components. For each case, evaluate whether Aaron's diagrams are correct or incorrect. If incorrect, explain the problem or make the correction.

2. Fill in the blanks

3. Fill in the blanks

Horizontally Launched Projectile Problems vertical direction a y = -g v iy = 0, v y = -g∙t d y = - ½ g∙t 2 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. Horizontal direction a x = 0 v ix = v i d x =v i ∙t

Example A pool ball leaves a 0.60-meter 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 Information Vertical Information v ix = 2.4 m/s a x = 0 m/s/s y = -0.60 m v iy = 0 m/s a y = -9.8 m/s/s vertical direction a y = -g v iy = 0, v y = -g∙t d y = - ½ g∙t 2 Horizontal direction a x = 0 v ix = v i d x =v i ∙t t = 0.350 s x = 0.84 m

Example A soccer ball is kicked horizontally off a 22.0-meter 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. Horizontal Information Vertical Information x = 35.0 m a x = 0 m/s/s v ix = ? y = -22.0 m v iy = 0 m/s a y = -9.8 m/s/s vertical direction a y = -g v iy = 0, v y = -g∙t d y = - ½ g∙t 2 Horizontal direction a x = 0 v ix = v i d x =v i ∙t t = 2.12 s v ix = 16.5 m/s

Example The path of a stunt car driven horizontally off a cliff is represented in the diagram below. After leaving the cliff, the car falls freely to point A in 0.50 second and to point B in 1.00 second. 1.Determine the magnitude of the horizontal component of the velocity of the car at point B. [Neglect friction.] 2.Determine the magnitude of the vertical velocity of the car at point A. 3.Calculate the magnitude of the vertical displacement, d y, of the car from point A to point B. [Neglect friction.]

example A cannon elevated at an angle of 35° to the horizontal fires a cannonball, which travels the path shown in the diagram. [Neglect air resistance and assume the ball lands at the same height above the ground from which it was launched.] If the ball lands 7.0 × 10 2 meters from the cannon 7.0 seconds after it was fired, 1.what is the horizontal component of its initial velocity? 2.what is the vertical component of its initial velocity?

Example A machine fired several projectiles at the same angle, θ, above the horizontal. Each projectile was fired with a different initial velocity, v i. The graph below represents the relationship between the magnitude of the initial vertical velocity, v iy, and the magnitude of the corresponding initial velocity, v i, of these projectiles. Calculate the magnitude of the initial horizontal velocity of the projectile, v ix, when the magnitude of its initial velocity, v i, was 40. meters per second. Answer Hint Reason Properties

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