1 / 25

Yay Math!

Yay Math!. By Eve, Eli, Friederike, Shirley, Jasper and Catherine. „THE INTE GREAT S“. Our Problem . A ball is thrown into the air with a certain velocity. It reaches a certain height and falls then back to earth.

olympia-daugherty
Download Presentation

Yay Math!

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Yay Math! By Eve, Eli, Friederike, Shirley, Jasper and Catherine „THE INTEGREATS“

  2. Our Problem A ball is thrown into the air with a certain velocity. It reaches a certain height and falls then back to earth. Does it take longer to reach its maximum height or to fall back to earth from the maximum height?

  3. WHAT DOES YOUR PHYSICAL INTUITION TELL YOU? Think about the situation and make a guess!

  4. Which is Faster, Going Up or Coming Down? For this problem, we need to… • Solve differential equations • Prove that a given equation for the height of the ball is correct • Prove that a given equation for the time that the ball takes to reach its maximum height is correct • Solve this Time-Equation for given values • Discover if the ball is faster leaving Earth or coming back down • Use an indirect method of determining which part of the ball’s trajectory is faster.

  5. Question 1 A ball with mass m is projected vertically upward from Earth’s surface with a positive initial velocity v0. We assume the forces acting on the ball are the force of gravity and a retarding force of air resistance with direction opposite to the direction of motion and with magnitude p|v(t)| where p is a positive constant and v(t) is the velocity of the ball at time t. In both the ascent and the descent, the total force acting on the ball is –pv-mg. (During ascent, v(t) is positive and the resistance acts upward.) So, by Newton’s Second Law, the equation of motion is mv’=-pv-mg

  6. How to solve Question 1 mv’=-pv-mg Solve this differential equation to show that the velocity is…

  7. - To solve we need to use the integral Separate the variables Now: Apply Substitution Method

  8. Substitute: Now: Plug in –pv-mg for u

  9. Now: Solve for K

  10. - When t=0, K becomes –pv-mg (K=-pvo-mg);v becomes vo Now: Plug the value for K into the equation

  11. Question 2 Show that the height of the ball until it hits the ground, is…

  12. How to solve Question 2

  13. Now: Solve for c Condition: t=0; y=0 Now: Plug c back into the equation

  14. The second equation is the same as the first one only factored.

  15. Question 3 Let t1 be the time that the ball takes to reach its maximum height. Show that…

  16. How to solve Question 3 Condition: At the maximum height the velocity will be zero, so we set v(t)=0

  17. The common denominator for both terms is p, so vo needs to be extended (multiplied) by p, in order to make this happen.

  18. http://www.blackgold.ab.ca/ict/Division4/Math/Div.%204/Ball%20Toss/index.htmhttp://www.blackgold.ab.ca/ict/Division4/Math/Div.%204/Ball%20Toss/index.htm Question 4: Graphing (ideal)

  19. Question 5 In general, it‘s not easy to find t2 (the time at which the ball falls back to earth) because it is impossible to solve the equation y(t)=0 explicitly. We can, however, use an indirect method to determine whether ascent or descent is faster; we determine whether y(2t1) is positive or negative.

  20. Plugging all values that we are given in, we can determine whether it is positive or negative.

  21. Actual graph: 2.07

More Related