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What is a projectile?

What is a projectile?. A projectile is object that is thrown and is un-powered. Examples of projectiles Launched Arrows Thrown footballs, baseballs,ect Car running off a cliff or ramp A projectile does only two things Falls with an acceleration of 9.8 m/s 2 (DOWNWARD)

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What is a projectile?

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  1. What is a projectile? • A projectile is object that is thrown and is un-powered. • Examples of projectiles • Launched Arrows • Thrown footballs, baseballs,ect • Car running off a cliff or ramp • A projectile does only two things • Falls with an acceleration of 9.8 m/s2(DOWNWARD) • Moves horizontally (sideways) with a constant velocity.

  2. Horizontal motion: The distance traveled every second is constant. The velocity is constant. Vertical motion:The distance traveled every second is increasing. The velocity is increasing. Properties of Projectile Motion • Independence of path. • How a projectile moves along the x- axis is independent with how it moves along the y-axis.

  3. Graphing Projectile Motion’s Velocity Slope (A.K.A. Acceleration)= 0 m/s/s VY VX Velocity Time Slope (A.K.A. Acceleration)= -9.8 m/s/s When graphing a projectile’s velocity with respect to time is it important to realize the that projectile really has two different velocities [one going across the ground (X axis), and one moving vertically (Y axis)]

  4. Graphing Projectile Motion’s Positions Constant slope means constant velocity (acceleration = 0 m/s/s) PositionX PostionY Position Time Just like when we graph a projectile’s velocity with respect to time we realize the that projectile really has two different velocities, a projectile also has two different positions[ the distance across the ground (X axis), and height (Y axis)] Upside down parabola means slope is not constant. Negative acceleration.

  5. E D F C G B H A I X up X down Xup = Xdown Properties of Projectile Motion The Acceleration is always 9.8m/s2 Downward!!!!! • Projectile motion is symmetrical around the “maximum height” of the projectile's path SpeedA = SpeedI SpeedB = SpeedH SpeedC = SpeedG SpeedD = SpeedF From A to E speed slows down Since Vx is constant and X up = X down Then Time up = Time down From E to I speed speeds up Speed is slowest at E

  6. Changes in a Projectile’s Velocity At max height Vy is ALWAYS 0 m/s!! The acceleration is, however, 9.8 m/s2 downward As a projectile reaches maximum height the velocity slows down and flattens out. Vy goes to 0 m/s at a rate of 9.8m/s2, Vx is a constant As the projectile falls from maximum height the velocity increases and becomes more vertical Vy increases at a rate of 9.8m/s2 VX is constant

  7. Range is how far along the ground the projectile travels Range Angles and Ranges If we launch several different projectiles at different angles but with the same speed, and there is NO AIR RESISTANCE we will see; as the angle launched increases from 0 degrees to 45 degrees the range increase, as the angle launched increases from 45 degrees to 90 degrees the range degreases. The range is a maximum when the launching angle is 45 degrees. Two different angles can give you the same range, when launched with the same speed. These two angles must add up to 90 degrees. A projectile launched with a speed of 100 m/s at an angle of 35 degrees has the same range as a projectile launched at a speed of 100 m/s launched at an angle of 55 degrees.

  8. Solving projectile problems • The trick is not to do too much at one time. A projectile has parts. • The X-axis motion which is a constant velocity • The Y-axis motion which is accelerated at 9.8m/s2 Break any projectile problem into these two parts, and solve for the time of the trip. Then find what you are asked for. The Key is finding time!!!!

  9. (0m, 4.9m) (range, 0m) Example problem 1 A ball rolls off a 4.9 meter tall table with A speed of 3m/s. What is the range of the ball? 3 m/s 4.9m Set up: 1st Get a Visual: Draw the situation 2nd Break the problem into components: Draw X & Y axis's and label the initial and final points of the projectile. (You can have the origin anywhere)

  10. 3 m/s (0m, 4.9m) (range, 0m) Example problem 1 A ball rolls off a 4.9 meter tall table with A speed of 3m/s. What is the range of the ball? X- axis Y axis ax = -9.8 m/s/s VYi = 0 m/s PosYi = 4.9 m PosYf = 0 m ax = 0 m/s/s Vxi = 3 m/s Posxi = 0 m Posxf = range = ? Set up (continued) : 3rd Organize the information: Set up a table as shown and record all the known info. IT IS IMPORTANT TO REMEMBER THAT WE ALREADY KNOW THE ACCELERATIONS FOR THE X AND Y AXISES (ax = 0 m/s/s, and ay = - 9.8 m/s/s)

  11. Step 1: Write out the Position equation twice (one for the X axis, one for the Y axis) 3 m/s (0m, 4.9m) (range, 0m) PosYf = PosYi + VYit +(½)ayt2 PosXf = PosXi + VXit + (½)aXt2 Step 2: Substitute the information into both equations 0m = 4.9 m + (0 ms/) t +(1/2)(-9.8 m/s2)t2 Range = 0m + (3m/s)t + (.5)(0m/s/s) t2 -4.9 m = (-4.9 m/s2)t2 Range = (3m/s)t Step 4: Solve for what you are asked to find 1 s2 = t2 Range = (3m/s)(1s) 1 s = t Range = 3m Step 3: Find the time Example problem 1 A ball rolls off a 4.9 meter tall table with A speed of 3 m/s. What is the range of the ball? X- axis Y axis ax = -9.8 m/s/s VYi = 0 m/s PosYi = 4.9 m PosYf = 0 m ax = 0 m/s/s Vxi = 3 m/s Posxi = 0 m Posxf = range = ?

  12. Example problem 2 A ball moving .5 m/s rolls off a table and has a range of .25 meters. How tall is the table? (0m, Height) .5 m/s (.25m, 0m) Height .25 m Set up: 1st Get a Visual: Draw the situation 2nd Break the problem into components: Draw X & Y axis's and label the initial and final points of the projectile. (You can have the origin anywhere)

  13. (0m, Height) .5 m/s (.25m, 0m) Height .25 m Example problem 2 A ball moving .5 m/s rolls off a table and has a range of .25 meters. How tall is the table? X- axis Y axis ax = -9.8 m/s/s VYi = 0 m/s PosYi = Height = ? PosYf = 0 m ax = 0 m/s/s Vxi = .5 m/s Posxi = 0 m Posxf = .25 m Set up (continued) : 3rd Organize the information: Set up a table as shown and record all the known info. IT IS IMPORTANT TO REMEMBER THAT WE ALREADY KNOW THE ACCELERATIONS FOR THE X AND Y AXISES (ax = 0 m/s/s, and ay = - 9.8 m/s/s)

  14. Step 1: Write out the Position equation twice (one for the X axis, one for the Y axis) 3 m/s (0m, 4.9m) (range, 0m) PosYf = PosYi + VYit +(½)ayt2 PosXf = PosXi + VXit + (½)aXt2 Step 2: Substitute the information into both equations 0m = Height + (0 ms/) t +(1/2)(-9.8 m/s2)t2 .25 m = 0m + (.5m/s)t + (.5)(0m/s/s) t2 -Height = (-4.9 m/s2)t2 .25 m = (.5 m/s)t Step 4: Solve for what you are asked to find +Height = (+4.9 m/s2)t2 .5 s = t +Height = (+4.9 m/s2)(.5 s)2 Height = 1.225 meters Step 3: Find the time Example problem 2 A ball rolls off a 4.9 meter tall table with A speed of 3 m/s. What is the range of the ball? X- axis Y axis ax = 0 m/s/s Vxi = .5 m/s Posxi = 0 m Posxf = .25 m ax = -9.8 m/s/s VYi = 0 m/s PosYi = Height = ? PosYf = 0 m

  15. Velocities at an Angle. The problems are worked pretty much the same way. The only diffre4nce is after you set up your X and Y axis's, you break the velocity vector into X and Y components

  16. Vi = 20 m/s @ 30O Range Example problem 3 A soccer ball is kicked from ground level with a speed of 20 m/s and an angle of 30 degrees. What is the ball’s range?

  17. Example problem 3 A soccer ball is kicked from ground level with a speed of 20 m/s and an angle of 30 degrees. What is the ball’s range? Vi = 20 m/s @ 30O (range, 0m) (0m, 0m) Range Set up: 1st Get a Visual: Draw the situation 2nd Break the problem into components: Draw X & Y axis's and label the initial and final points of the projectile. (You can have the origin anywhere)

  18. VYi = (20 m/s)* Sin (30O) VYi = 10 m/s VXi = (20 m/s)*Cos (30O) VXi = 17.32 m/s Use Sine and Cosine to break the original velocity vector into its X and Y components. Example problem 3 A soccer ball is kicked from ground level with a speed of 20 m/s and an angle of 30 degrees. What is the ball’s range? 20 m/s 30O

  19. Vi = 20 m/s @ 30O (range, 0m) (0m, 0m) Range Example problem 3 A soccer ball is kicked from ground level with a speed of 20 m/s and an angle of 30 degrees. What is the ball’s range? X- axis Y axis ax = -9.8 m/s/s VYi = 10 m/s PosYi = 0 m PosYf = 0 m ax = 0 m/s/s Vxi = 17.32 m/s Posxi = 0 m Posxf = range = ? Set up (continued) : 3rd Organize the information: Set up a table as shown and record all the known info. IT IS IMPORTANT TO REMEMBER THAT WE ALREADY KNOW THE ACCELERATIONS FOR THE X AND Y AXISES (ax = 0 m/s/s, and ay = - 9.8 m/s/s)

  20. Step 1: Write out the Position equation twice (one for the X axis, one for the Y axis) PosYf = PosYi + VYit +(½)ayt2 PosXf = PosXi + VXit + (½)aXt2 0m = 0 m + (10 ms/) t +(1/2)(-9.8 m/s2)t2 Range = 0m + (17.32 m/s)t + (.5)(0m/s/s)t2 0 m = (10 m/s)t + (-4.9 m/s2)t2 Range = (17.32 m/s)t Step 4: Solve for what you are asked to find - (10 m/s)t = (-4.9 m/s2)t2 Time is on both sides of the equation so one t cancels out on each side. Range = (17.32m/s)(2.04 s) -(10 m/s)/(-4.9 m/s2)= t Step 3: Find the time Range = 35.33 m 2.04 s = t Step 2: Substitute the information into both equations A soccer ball is kicked from ground level with a speed of 20 m/s and an angle of 30 degrees. What is the ball’s range? Vi = 20 m/s @ 30O Example problem 3 (range, 0m) Range (0m, 0m) X- axis Y axis ax = -9.8 m/s/s VYi = 10 m/s PosYi = 0 m PosYf = 0 m ax = 0 m/s/s Vxi = 17.32 m/s Posxi = 0 m Posxf = range = ?

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