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Physics Honors

Physics Honors. Good News/Bad News: These are the same formulas we used for linear motion. Do you know them?. If the answer is “NO”, then memorize them NOW!. We assume NO AIR RESISTANCE! (Welcome to “Newtonia”), therefore… The path of a projectile is a parabola.

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Physics Honors

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  1. Physics Honors

  2. Good News/Bad News: These are the same formulas we used for linear motion. Do you know them? If the answer is “NO”, then memorize them NOW!

  3. We assume NO AIR RESISTANCE! (Welcome to “Newtonia”), 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.

  4. 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 “a = - 9.81 m/s2”.

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

  6. A cannon ball is shot horizontally from a cliff. vx Knowns: Δdy Range, Δdx Remember to keep the horizontal and vertical motions separate. Time is the factor that ties them together. Since the vertical motion is treated like “free-fall” and we have more info about the vertical, we should use that to find time first.

  7. 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: Δdy= -35 m Range, Δ dx Add the given values to our list of known values. Now that the diagram is drawn and labeled and we have identified and listed all of our “known” and “given” values for the problem, let’s begin by finding time.

  8. A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ? Vx= 5 m/s Knowns: Givens: Δdy= -35 m Range, Δdx Since we know more values for vertical motion, let’s use it to find time. Start with the distance equation… Now solve for t…

  9. A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, Δdx = ? Find final velocity, vf = ? Vx= 5 m/s Knowns: Givens: Δdy= -35 m Calculated: Range, Δdx Now that we know time, let’s find dx. Remember that horizontal motion is constant. Let’s use the distance formula again. This time in the x – direction… 0

  10. A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, Δdx = ? Find final velocity, vf = ? Vx= 5 m/s Knowns: Givens: Δdy= -35 m Vf Calculated: Range, Δdx Final velocity requires a little more thought. Remember that 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.

  11. A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, Δdx = ? Find final velocity, vf = ? Vx= 5 m/s Knowns: Givens: Δdy= -35 m Vf Calculated: Range, Δ dx θ Let’s look more closely at the vector, vf. To help see it better, let’s exaggerate the angle. Since x- and y- motion are separate, there must be components. So, we have the x-component already due to the fact that horizontal velocity is constant. Before we can find vf, we must find the vertical component, vfy. Vf Vfy θ Vfx = 5 m/s

  12. A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, Δdx = ? Find final velocity, vf = ? Vx= 5 m/s Knowns: Givens: Δdy= -35 m Vf Calculated: Range,Δdx θ 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. Calculating vfy: Vf Vfy θ Vfx = 5 m/s Still not finished. Gotta put components together for final.

  13. A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, Δdx = ? Find final velocity, vf = ? Vx= 5 m/s Knowns: Givens: Δdy= -35 m Vf Calculated: Range,Δdx θ 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. Putting it together to calculate vf: Vf θ Vfx = 5 m/s Final Velocity = 26.7m/s@280.2°

  14. v θ voy vox 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 “v” 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. **Never use the original velocity at the angle in an equation! v voy θ vox

  15. vf vi θ θ vfx vf vfy θ vfx vfy voy vox 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…

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

  17. Vy top = 0 Voy dymax =height Keeping the horizontal and vertical motions separate! v Vx θ Range, ∆dx Knowns (for all projectiles): Step 2: Divide initial velocity into horizontal (x) and vertical (y) components. Step 3: Find time if possible. Use vertical motion.

  18. 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.

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

  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 V= 30 m/s Voy 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.

  21. Vy top = 0 voy=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. v= 30 m/s 60° vx=15 m/s Range, ∆dx Finding time – Method 1: 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.

  22. 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 Voy vfy dymax =height v 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:

  23. 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 voy=26 m/s vy =-26 m/s θ dymax =height v 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

  24. 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 voy=26 m/s vy =-26 m/s θ dymax =height v vx =15 m/s vx =15 m/s θ Range, ∆dx To find maximum height for this problem, remember that because it is launched and lands at the same level, maximum height occurs exactly half-way through the flight. So…USE ½ of the total time of flight.

  25. Projectiles that land higher than they are launched. vi dy θ dx Vertical displacement is not zero. Consider the launch point as the zero height and then vertical displacement, dy, will be a positive number (as long as you continue to assume up is positive. Plug the value for dy into the above equation and solve for time. (Hint: Graph it and find the zeros! It’s easier than the quadratic equation.) The answer must be positive. This will be the time for the flight to that point. Then you can use that time to find the horizontal distance the object traveled to get to that point. 0 NOTE: The highest point of the projectile DOES NOT occur at the half-way point of the flight. BE CAREFUL!

  26. Projectiles that land lower than they are launched. Again, vertical displacement is not zero. If the launch point as the zero height, vertical displacement, dy, will be a negative number (as long as you continue to assume up is positive. Plug the value for dy into the distance equation and solve for time. This will be the time for the flight to that point. Then you can use that time to find the horizontal distance the object traveled to get to that point. NOTE: The highest point of the projectile DOES NOT occur at the half-way point of the flight. BE CAREFUL! vi θ -dy dx 0

  27. Hints for projectiles that are thrown downward The key to this problem is drawing the picture and finding the components. Again, vertical displacement is negative along with the vertical component of the initial velocity and acceleration. Remember, to find final velocity, you must find the final y- velocity and then put the x- and y- components of the final velocity back together using the pythagorean theorem and the inverse tangent. vix θi vi viy vfx -dy vf vfy θf dx Some important notes: The horizontal velocity remains the same (drawing shows otherwise but it is not to scale). The angle is is thrown downward is not the same as the angle it strikes the ground. Finally, the projectile will never strike perpendicular to the ground! 0

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