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Projectile Motion Horizontally Launched Projectiles Projectiles Launched at an Angle A.S. 2.1.11 – 2.1.14 Due Friday, 11/14 Text Reference: chapter 3.

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Presentation on theme: "Projectile Motion Horizontally Launched Projectiles Projectiles Launched at an Angle A.S. 2.1.11 – 2.1.14 Due Friday, 11/14 Text Reference: chapter 3."— Presentation transcript:

1 Projectile Motion Horizontally Launched Projectiles Projectiles Launched at an Angle A.S. 2.1.11 – 2.1.14 Due Friday, 11/14 Text Reference: chapter 3 Projectile Motion Horizontally Launched Projectiles Projectiles Launched at an Angle A.S.

2 Question 3.4a Firing Balls I A small cart is rolling at constant velocity on a flat track. It fires a ball straight up into the air as it moves. After it is fired, what happens to the ball? a) it depends on how fast the cart is moving b) it falls behind the cart c) it falls in front of the cart d) it falls right back into the cart e) it remains at rest Question 3.4a Firing Balls I A small cart is rolling at constant velocity on a flat track.

3 Now the cart is being pulled along a horizontal track by an external force (a weight hanging over the table edge) and accelerating. It fires a ball straight out of the cannon as it moves. After it is fired, what happens to the ball? a) it depends upon how much the track is tilted b) it falls behind the cart c) it falls in front of the cart d) it falls right back into the cart e) it remains at rest Question 3.4b Firing Balls II Now the cart is being pulled along a horizontal track by an external force (a weight hanging over the table edge) and accelerating.

4 The same small cart is now rolling down an inclined track and accelerating. It fires a ball straight out of the cannon as it moves. After it is fired, what happens to the ball? a) it depends upon how much the track is tilted b) it falls behind the cart c) it falls in front of the cart d) it falls right back into the cart e) it remains at rest Question 3.4c Firing Balls III The same small cart is now rolling down an inclined track and accelerating.

5 Question 3.6aDropping the Ball I From the same height (and at the same time), one ball is dropped and another ball is fired horizontally. Which one will hit the ground first? a) the “dropped” ball b) the “fired” ball c) they both hit at the same time d) it depends on how hard the ball was fired e) it depends on the initial height Question 3.6aDropping the Ball I From the same height (and at the same time), one ball is dropped and another ball is fired horizontally.

6 Question 3.6bDropping the Ball II In the previous problem, which ball has the greater velocity at ground level? a) the “dropped” ball b) the “fired” ball c) neither—they both have the same velocity on impact d) it depends on how hard the ball was thrown Question 3.6bDropping the Ball II In the previous problem, which ball has the greater velocity at ground level.

7 Components of Motion--horizontal Horizontal motion is unaffected by vertical accelerations or vertical motion We will work with the assumption that air resistance is negligible, so… Horizontal velocity will remain CONSTANT throughout the flight of a projectile Components of Motion--horizontal Horizontal motion is unaffected by vertical accelerations or vertical motion We will work with the assumption that air resistance is negligible, so… Horizontal velocity will remain CONSTANT throughout the flight of a projectile

8 Horizontally Launched: Problem Solving StrategyX-directionY-direction v =u = s =v = t =a = s = t = Horizontally Launched: Problem Solving StrategyX-directionY-direction v =u = s =v = t =a = s = t =

9 Practice Problem A marble rolls across a horizontal table with a speed of 2.50 m·s -1. If the table is 1.12 m high, how far away from the edge of the table will the marble land? X-directionY-direction v = 2.50 m·s -1 u = 0 s = ?v = t = ?a = -9.81 m·s -2 s = 1.12 m t = ? Step 1: Solve for time in the air. Which direction gives you enough information to find the time? Step 2: Calculate the horizontal displacement from the calculated time and the given velocity. Practice Problem A marble rolls across a horizontal table with a speed of 2.50 m·s -1.

10 X-directionY-direction v = 2.50 m·s -1 u = 0 s = ?v = t = ?a = -9.81 m·s -2 s = 1.12 m t = ? X-directionY-direction v = 2.50 m·s -1 u = 0 s = v = t = a = m·s -2 s = 1.12 m t =

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12 General Terms/Characteristics Range: the horizontal displacement of a projectile Trajectory: The path the projectile follows while in the air Flight Time (or, Hang Time): The total amount of time the projectile is in the air. Maximum Height: The vertical displacement a projectile has reached when its vertical velocity is zero. General Terms/Characteristics Range: the horizontal displacement of a projectile Trajectory: The path the projectile follows while in the air Flight Time (or, Hang Time): The total amount of time the projectile is in the air.

13 Projectile motion can only be analyzed by breaking the initial velocity into its horizontal and vertical components: Just like with horizontal launches, the horizontal velocity will remain constant, while the vertical motion will be affected by gravity Velocity Components Projectile motion can only be analyzed by breaking the initial velocity into its horizontal and vertical components: Just like with horizontal launches, the horizontal velocity will remain constant, while the vertical motion will be affected by gravity Velocity Components

14 Problem Solving Technique: Just like with horizontally launched projectiles, you must separate all variables into the appropriate direction: X-directionY-direction Vx = d x = t = V iy = a = d y = t = X-directionY-direction Vx = v·cos(  ) d x = range t = flight time V iy = v·sin(  ) a = - 9.81 m/s 2 d y = Max. height t = ½ flight time Problem Solving Technique: Just like with horizontally launched projectiles, you must separate all variables into the appropriate direction: X-directionY-direction Vx = d x = t = V iy = a = d y = t = X-directionY-direction Vx = v·cos(  ) d x = range t = flight time V iy = v·sin(  ) a = m/s 2 d y = Max.

15 Example Problem #1: A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the horizontal. Find the ball’s: hang time Range maximum height Example Problem #1: A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the horizontal.

16 Example #2 An arrow is shot at 30.0° above the horizontal. Its velocity is 49.0 m/s and it hits the target. What is the maximum height the arrow will attain? The target is at the same height from which the arrow was originally shot. How far away is it? Example #2 An arrow is shot at 30.0° above the horizontal.

17 Example #3 A soccer ball is kicked into the air at a velocity of 29.0 m/s and at an angle of 37.0°. How far away will it first hit the ground? (Assuming we ignore air resistance) Example #3 A soccer ball is kicked into the air at a velocity of 29.0 m/s and at an angle of 37.0°.

18 Range Equation The kinematic equations for the vertical and the horizontal directions can be combined mathematically in order to solve for the horizontal range WITHOUT KNOWING THE TIME IN THE AIR! R = 2v i 2 ·Sin  ·Cos  g R = range (m) V i = overall launch velocity (the total vector quantity) (m/s) g = 9.81 m/s 2  *Note: this is NOT negative! Range Equation The kinematic equations for the vertical and the horizontal directions can be combined mathematically in order to solve for the horizontal range WITHOUT KNOWING THE TIME IN THE AIR.

19 Example #3 A pitched ball is hit by a batter at a 45° angle and just clears the outfield fence, 98 m away. Assuming that the fence is at the same height as the pitch, find the velocity of the ball it left the bat. Example #3 A pitched ball is hit by a batter at a 45° angle and just clears the outfield fence, 98 m away.


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