Yay Math!Yay Math!

By Eve, Eli, Friederike, Shirley, By Eve, Eli, Friederike, Shirley, Jasper and CatherineJasper and Catherine

„THE INTEGREATS“

Our ProblemOur Problem

A ball is thrown into the air with a A ball is thrown into the air with a certain velocity. It reaches a certain certain velocity. It reaches a certain height and falls then back to earth. height and falls then back to earth.

Does it take longer to reach its Does it take longer to reach its maximum height or to fall back to maximum height or to fall back to earth from the maximum height?earth from the maximum height?

WHAT DOES YOUR WHAT DOES YOUR PHYSICAL PHYSICAL INTUITION TELL INTUITION TELL YOU?YOU?

Think about the situation and Think about the situation and make make a guess!a guess!

Which is Faster, Going Up or Which is Faster, Going Up or Coming Down?Coming Down?

For this problem, we need to…For this problem, we need to…– Solve differential equations Solve differential equations – Prove that a given equation for the height of the Prove that a given equation for the height of the

ball is correctball is correct– Prove that a given equation for the time that the Prove that a given equation for the time that the

ball takes to reach its maximum height is correctball takes to reach its maximum height is correct– Solve this Time-Equation for given valuesSolve this Time-Equation for given values– Discover if the ball is faster leaving Earth or Discover if the ball is faster leaving Earth or

coming back down coming back down – Use an indirect method of determining which Use an indirect method of determining which

part of the ball’s trajectory is faster. part of the ball’s trajectory is faster.

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

mv’=-pv-mgmv’=-pv-mg

How to solve Question 1How to solve Question 1

mv’=-pv-mgmv’=-pv-mg

Solve this differential equation to show that the Solve this differential equation to show that the velocity is…velocity is…

v(t) (vomg

p)e pt /m

mg

p

- To solve we need to use the integral- To solve we need to use the integral

Separate the Separate the variablesvariables

Now: Apply Now: Apply Substitution MethodSubstitution Method

v'dv

dt

mdv

dt pv mg

mdv

pv mgdt

mdv

pv mg dt

mdv

pv mg dt

Substitute:Substitute:

Now: Plug in Now: Plug in –pv-mg –pv-mg for ufor u

ctup

m

dtduup

m

dtu

du

p

m

dtu

dvdvdu

p

m

dtmgpv

pdv

p

m

)ln(

1

pdv

du

mgpvu

Now: Solve for K

- When t=0, K becomes –pv-mg (K=-pvo-mg);v becomes vo

Now: Plug the value for K into the equation

QuestionQuestion 2 2

Show that the height of the Show that the height of the ball until it hits the ground, ball until it hits the ground,

is…is…

How to solve Question 2How to solve Question 2

Now: Solve for c Condition: t=0; y=0

Now: Plug c back into the equation

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

Question 3Question 3

Let tLet t1 1 be the time that the ball be the time that the ball takes to reach its maximum takes to reach its maximum height. Show that…height. Show that…

How to solve Question 3How to solve Question 3

At the maximum height the velocity will be zero, so we set v(t)=0

Condition:

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

http://www.blackgold.ab.ca/ict/Division4/Math/Div.%204/Ball%20Toss/index.htm

Question 4: Graphing (ideal)

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

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

Actual graph:

2.07