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Chapter 3 Projectile motion Kinematics in Dimensions Two

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Projectile Motion Projectile motion is motion in two directions Motion in the x-direction is independent of the y-direction Motion in the y-direction is independent of the x-direction

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What is a Projectile? A projectile is any object that is placed into free flight and is being affected by gravity. The path of a projectile is called the trajectory.

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Throw The Full Parabola VyVy VxVx The key to the full parabola is symmetry. Try to identify some points of symmetry.

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Half Parabola Timing The time of flight of a half parabolic path is equal to that of simply dropping the object from the same height. Horizontal velocity (v x ) has no affect on flight time because it is not affected by gravity.

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Jill drops the yellow ball and throws the red ball horizontally. Which ball will hit the ground first?

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X & Y are Independent

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Object launched horizontally Multi media studio Horizontally launched projectiles The plane and the package The truck and the ball What do you notice about the horizontal velocity in each of the following animations?

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A stone is thrown horizontally from the top of a 78.4m high cliff at 5m/s. a) How long does it take to reach the bottom? VarXYWant a v1v1 v2v2 d1d1 d2d2 t

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A stone is thrown horizontally from the top of a 78.4m high cliff at 5m/s. a) How long does it take to reach the bottom?

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A stone is thrown horizontally from the top of a 78.4m high cliff at 5m/s. b) How far from the base does it land?

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A stone is thrown horizontally from the top of a 78.4m high cliff at 5m/s. c) What are the final v y and v x ?

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Sample Partial WS 7a #1 A cannon nestled in the side of a cliff (d 1y = 65m) fires a cannon ball at 26. How long until the ball splashes into the sea? Fire

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Sample Problem A toy car is raced off a table (1.1m high) onto the floor below. –How long did it take for the car to crash on the floor? 0

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Projectile Motion Type Not all object are launch horizontally Objects can be launched at an angle

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Recall the trajectory of the golf ball when hit with a 3 iron. What would the trajectory of a 9 iron look like? The loft of the club changed the launch angle.

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Object 1 was launched at 60 o Object 2 was launched at 30 o

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Object 1 was launched from a 25m high cliff at 0 o Object 2 was launched at 60 o

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Initial Velocity Breakdown When an object is launched at some angle, it’s initial velocity (v 1 ) can be broken down into two components. –Horizontal Component (V x ) –Vertical Component (V y ) What shape is formed? Consider also the launch angle ( ). v1v1 v 1y v 1x Please Note: horizontal and vertical components are independent of one another. The only commonality is time.

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Initial Velocity Breakdown (Cont.) Consider the breakdown from the previous slide again. There are trigonometric relationships between the sides and angles of a right triangle. v1v1 v 1y v 1x Important!

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Practicing Trig Functions Consider the triangle below. Solve for the unknown values. 13 x y 22.62° Searching for xSearching for y

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Sample Velocity Breakdown A dart gun is fired at an angle of 30° with a muzzle velocity of 40m/s. Calculate the components of the velocity? v1v1 v 1y v 1x Horizontal Component (x)Vertical Component (y)

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Given an initial velocity of 40m/s and an angle of 25 find v 1x & v 1y V 1 =40 v 1y v 1x =25 o Searching for x Searching for y

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Sample Full Parabola Problem A golf ball is struck at an angle of = 36° with the horizontal at a velocity of 45m/s. –What are the components of the velocity (v 1x and v 1y )? Strike v 1x v 1y v1v1 Horizontal Component Vertical Component

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Problem Solving Strategies Solve for the horizontal component V xi –Use trig functions Solve for the vertical component V yi –Use trig functions Solve each direction (x & y) separately Symmetry can be used when the launching & landing places are the same height.

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a)Find the hang time b)find the horizontal distance the ball travels. b)The maximum height of the ball. A football player kicks a ball at 27m/s at an angle of 30°. HorizontalVertical VKWVKW ag-9.8 V 1x v 1y v 2x v 2y dxdx dxdx dydy dydy ttt

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Problem Solving Strategies Step 1: Solve for the horizontal and vertical components (V 1x & V 1y ) V=27m/s V 1x = ?m/s V 1y = ?m/s Searching for x Searching for y

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Problem Solving Strategies Step 2: Solve each direction (x & y) separately Symmetry can be used when the launching & landing places are the same height. 12.5m/s 10.0m/s 7.50m/s 5.00m/s 0.00m/s 15.0m/s 2.50m/s

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a)Find the hang time b)find the horizontal distance the ball travels. b)The maximum height of the ball. A football player kicks a ball at 27m/s at an angle of 30°. HorizontalVertical VKWVKW ag-9.8 v 1x 23.4V 1y 13.5 v 2x 23.4v2yv2y dxdx dxdx dydy dydy ttt

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a)Find the hang time A football player kicks a ball at 27m/s at an angle of 30°.

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b) Find the horizontal distance A football player kicks a ball at 27m/s at an angle of 30°.

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12.5m/s 10.0m/s 7.50m/s 5.00m/s 0.00m/s 15.0m/s 2.50m/s A football player kicks a ball at 27m/s at an angle of 30°. c) Find the maximum height What is true about the vertical velocity at the maximum height?

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A football player kicks a ball at 27m/s at an angle of 30°. Find the max height

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An arrow is shot at 44m/s at an angle of 60° Find the maximum height of the arrow. Find the horizontal distance the arrows travels. Find the hang time

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Problem Solving Strategies Step 1: Solve for the horizontal and vertical components (V 1x & V 1y ) V=44m/s V 1x = ?m/s V 1y = ?m/s Searching for x Searching for y

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a)Find the hang time b)find the horizontal distance the arrows travels. c) The maximum height of the arrow. An arrow is shot at 44m/s at an angle of 60° HorizontalVertical VKWVKW ag-9.8 v 1x 22v 1y 38.1 v 2x v 2y dxdx dxdx dydy dydy ttt

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a)Find the hang time A football player kicks a ball at 44m/s at an angle of 60°.

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b) Find the horizontal distance A football player kicks a ball at 44m/s at an angle of 60°.

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A football player kicks a ball at 44m/s at an angle of 60°. Find the max height Recall, v y =0 at d y max

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Adding Vectors Graphically You walk 5m @ 0 o and then turns to walk 6m @90 o. Finally, you turn to walk 8 m at 200°. What is your displacement? Addition is commutative!

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Adding Force Vectors Analytically hypotenuse adjacent opposite

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Components of Vectors Finding the vector magnitude and direction when you know the components. Recall: is measured from the positive x axis. Caution: Beware of the tangent function. Always consider in which quadrant the vector lies when dealing with the tangent function.

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8.66 5 -8.66 5 -5 8.66 -5

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Adding Vectors Analytically MagnitudeAngleX componentY component 4.5N30 o 3.89N2.25N 7N210 o -6.06N-3.5N 6N150 o -5.19N3.0N ---------------------------R x =-7.36NR y =1.7N

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Adding Vectors Analytically MagnitudeAngleX componentY component 4.5N30 o 3.89N2.25N 7N210 o -6.06N-3.5N 6N150 o -5.19N3.0N ---------------------------R x =-7.36NR y =1.7N R=7.55NAngle =-13 o +180 o = 167 o

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Concept Questions A stone is thrown horizontally from a cliff. How would the x distance change if the stone was thrown twice as fast? d x distance would double

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Concept Questions How would the v 2y change if the stone was thrown twice as fast? V x and V y are independent Since it was thrown horizontally, it would not change.

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Concept Questions How would the v 2y change if the cliff was twice as high?

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The Partial Parabola Recall, this path has elevation and launch angle. The trajectory again has an apex. This is mathematically the most complex path. Fire

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Sample Partial Parabola Problem A cannon nestled in the side of a cliff (d 1y = 20m) fires a cannonball at 130m/s at a 40° angle. –What are the components of the initial velocity? Fire Horizontal ComponentVertical Component

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Sample Partial Parabola Problem A cannon nestled in the side of a cliff (d 1y = 20m) fires a cannonball at 130m/s at a 40° angle. Fire

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You are trying to win a prize by throwing an apple into a basket on top of a pedestal. The apple leaves your hand 1.00 m beneath the top of the pedestal. The apple flies 3.09 m horizontally before landing in the bottom of the basket. The apple’s maximum height was 1.26 m. What was the apple’s initial velocity (magnitude and direction)? Partial Parabola Variablexy a v1v1 v2v2 d1d1 d2d2 t

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Partial Parabola Find V 1y knowing D y max = 1.26m D y (max) = 1.26m

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Partial Parabola Find the hang time knowing v 1y and ending height d 2y =1m

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Partial Parabola Partial parabolas can be represented by two half parabolas Total hang time is the time of the ½ parabola on the way up plus the time of the ½ parabola on the way down.

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The Partial Parabola If you look at this path carefully, you can see two half parabolas, which simplifies things considerably. You still must consider the launch angle and the components of the velocity when trying to solve.

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Mortar Problems A mortar crew fires a projectile at an enemy ammunitions storage facility that is protected by a wall located on top of a 200.0 m high cliff. The ammunition is located a horizontal distance of 314.68 m from the mortar’s position. The projectile passes directly over the wall at its maximum height of 215.24 m. What was the projectile’s initial velocity (magnitude and direction)? Vxy a v1v1 v2v2 d1d1 d2d2 t

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© 2001-2005 Shannon W. Helzer. All Rights Reserved. 1 Unit 4 Two Dimensional Kinematics Projectile Motion.

© 2001-2005 Shannon W. Helzer. All Rights Reserved. 1 Unit 4 Two Dimensional Kinematics Projectile Motion.

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