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Projectile Motion

Projectile Motion. Physics pre-AP. Equations of motion :. Concepts concerning Projectile Motion:. We assume NO AIR RESISTANCE! (Welcome to “Physicsland”), therefore… The path of a projectile is a parabola. Horizontal motion is constant velocity. Vertical motion is in “free-fall”.

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Projectile Motion

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  1. Projectile Motion Physics pre-AP

  2. Equations of motion:

  3. Concepts concerning Projectile Motion: • We assume NO AIR RESISTANCE! (Welcome to “Physicsland”), therefore… • The path of a projectile is a parabola. • Horizontal motion is constant velocity. • Vertical motion is in “free-fall”. • Vertical velocity at the top of the path is zero • Time is the same for both horizontal and vertical motions. • The horizontal and vertical motions are independent.

  4. How are the formulas different for projectiles? They must be applied along only one axis at a time. horizontal or “x” – direction vertical or “y” – direction 0 0 0 Remember that for projectiles, the horizontal and vertical motions must be separated and analyzed independently. Remember that “ax” is zero and “ay” is acceleration due to gravity “g = 9.81 m/s2”.

  5. Projectiles Launched Horizontally

  6. A typical physics problem A cannon ball is shot horizontally from a cliff. vx What do we know? For all projectiles… height dy vx Range, dx Hint: You should always list your known values at the beginning of any problem and assign those values variables.

  7. Where do we start? Knowns: vx height dy Range, dx Remember to keep the horizontal and vertical motions separate. Time is the same for both directions. The time to fall (flight time) depends on the height from which the projectile falls.

  8. Let’s put numbers on it. What do we want to find? A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time, t = ? Find range, dx = ? Find final velocity, vf = ? Vx= 5 m/s Knowns: Givens: dy=35 m height Range, dx Let’s begin by finding time.

  9. Finding time of flight for a horizontally launched projectile. Start with the distance equation… Vx= 5 m/s dy=35 m height Range, dx Now solve for t…

  10. Finding range (horizontal distance) Now that we know time, let’s find dx (the range). Remember that horizontal motion is constant so there is no acceleration. Let’s use the distance formula again. This time in the x – direction… Vx= 5 m/s dy=35 m height 0 Range, dx Notice: the distance in the horizontal direction is the horizontal velocity times the time. dx = vx · t

  11. Determining the velocity at impact: there are two components to the final velocity vector. The constant vx and the accelerated vy. Calculated: Vx= 5 m/s dy=35 m Final velocity is a vector quantity so we must state our answer as a magnitude (speed that the projectile strikes the ground) and direction (angle the projectile strikes the ground). Also remember horizontal velocity is constant, therefore the projectile will never strike the ground exactly at 90°. That means we need to look at the horizontal (x) and vertical (y) components that make up the final velocity. height Range, dx Vf

  12. So, we have the x-component already due to the fact that horizontal velocity is constant. Before we can find vimpact, we must find the vertical component, vfy. Vx= 5 m/s Calculated: dy=35 m height Vfx = 5 m/s (constant) Range, dx θ θ Calculating vfy: To find vfy, remember that vertical motion is in “free-fall” so it is accelerated by gravity from zero to some value just before it hits the ground. vimpact Vfy

  13. Putting everything together: Putting it together to calculate vf: Vx= 5 m/s dy=35 m height Range, dx θ Final Velocity = 26.7 m/s, 79.2° Now we know both the x- and y- components of the final velocity vector. We need to put them together for magnitude and direction of final velocity.

  14. Projectiles Launched at an Angle

  15. Separating the horizontal and vertical motions. vi viy θ vix Since the initial velocity represents motion in both the horizontal (x) and vertical (y) directions at the same time, we cannot use it in any of our equations. Remember, the most important thing about projectiles is that we must treat the horizontal (x) and vertical (y) completely separate from each other. So…we need to separate “vi” into its x- and y- components. We will use the method we used for vectors. Now that we have the components of the initial velocity, we will use only those for calculations. vi viy θ vix

  16. Symmetry of the projectiles path. vi viy vfy vf vf θ vix vfx θ θ A projectile’s path is a parabola, ALWAYS. That means if a projectile is launched and lands at the same height, there will be symmetry. The angle of launch and angle of landing will be equal. The initial velocity and the final velocity will be the same magnitude. Also, that means the components will be the same. Since the vertical motion is the same as a ball that is thrown straight up or dropped straight down (in free-fall), the y-components are equal and opposite. Since the horizontal motion is always at constant velocity… vfy vfx

  17. What do we know? Vy top = 0 Viy dymax =height vi Vx θ Range, dx Knowns (for all projectiles): Step 1: List known values! Draw and label picture.

  18. Vy top = 0 Vy dymax =height Keeping the horizontal and vertical motions separate! vi Vx θ Range, dx Step 2: Divide initial velocity into horizontal (x) and vertical (y) components. Step 3: Find flight time if possible. Use vertical motion.

  19. Vy top = 0 Vy dymax =height Finding time for projectiles that start and end at the same height. (Using distance formula.) Almost every projectile problem can be solved by starting with the displacement equation to solve for time. In this case… Note: There are three ways to find time for this problem. You may use any of them you wish. vi θ Range, dx Finding time – Method 1: Since the initial and final vertical positions are both the same, vertical displacement dy = 0. Now solve for time. Yes, it is a quadratic equation! This will be the time for the entire flight. NOTE: If you want to find maximum height you will only use half the time. If you want to find range, use the total time.

  20. Let's do one with numbers. A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height. Vy top = 0 Vi= 30 m/s Viy dymax =height Vx 60° Range, dx Knowns (for all projectiles): Given values: Step 1: List known values! Draw and label picture.

  21. Let's do one with numbers. A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height. Vy top = 0 Vi= 30 m/s Viy dymax =height Vx 60° Range, dx Step 2: Divide initial velocity into x- & y- components. Knowns (for all projectiles): Given values: We can add these to what we know. WE WILL NOT USE THE 30 m/s again in this problem because it is not purely an x- or y- value.

  22. Vy top = 0 vy =26 m/s dymax =height Finding time for projectiles that start and end at the same height. (Using distance formula.) A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height. vi= 30 m/s 60° vix =15 m/s Range, dx Finding time – Remember that dy = 0 because the projectile is starting and ending at the same level (y-position). So, using the known and given values for this problem and the components we calculated, we can solve for time.

  23. Let's review what we know before we go on… A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height. Vy top = 0 vf θ Vy vfy dymax =height vi Vx vfx θ Range, dx Knowns (for all projectiles): Givens: Now that I know time, I can add it to my list of known, given, and calculated values. To review… Calculated values:

  24. Finding range (horizontal distance) A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height. Vy top = 0 vf vy =26 m/s vy =-26 m/s θ dymax =height vi vx =15 m/s vx =15 m/s θ Range, dx Now that we know time, let’s calculate horizontal distance. Remember that horizontal acceleration is zero. 0

  25. Finding maximum height (vertical distance) A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height. Vy top = 0 vf vy =26 m/s vy =-26 m/s θ dymax =height vi vx =15 m/s vx =15 m/s θ Range, dx To find maximum height for this problem, remember that vytop = 0 So:

  26. Projectiles that do not land at the same height: vi dy θ dx Vertical displacement is not zero. Consider the launch point as the zero height (reference point) If you know the range: use range to find the time to that position. Δx = vx t Use this time in the distance equation to find the height at that position. NOTE: The highest point of the projectile DOES NOT occur at the half-way point of the flight. BE CAREFUL!

  27. Again, if the range is known use the method in the previous slide. If the range is not known, use quadratic formula to find the time to the landing. Then use this time to find the range. NOTE: The highest point of the projectile DOES NOT occur at the half-way point of the flight. BE CAREFUL! vi θ -dy dx

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