MTH 122 CSUGC The Choice of Quadratic Equation Discussion Responses

Please respond:POST1:The quadratic equation is used to describe the relationship between  objects that are thrown and how long it takes to hit the ground.In spirit of playoffs, I decided to determine how long a punt takes  to hit the ground. As I discovered in combing through ways of finding  relevant information, the science of a punt is pretty fascinating. For  instance, using a right triangle you can determine the necessary angle  for a kick to account for distance. That is a function by itself. For  this discussion I will just focus on the quadratic equation to determine  how long the average punt takes to hit the ground. This would be  relevant to know because members of a special team should be able to  cover this distance before the ball hits the return man.The formula_y=-4.9t^2+V0(t)+H0 where -4.9t^2 is the constant of gravity, V0 is  the initial velocity in Meters/Second and H0 is the starting height.This formula is created due to the effect of the football traveling  in a parabola. When the ball is first kicked it travels upward. Starting  from the initial launch, the effect of gravity begins to counteract the  initial velocity, eventually causing the football to fall. The actual  function output is height.Y=-4.9t^2+39.2t+2.45Solving by factoring: I don’t like this method. It usually doesn’t work to solve complex equations.Solving by completing the square_y=-4.9t^2+39.2t+2.45-4.9t^2+39.2t+2.45=0-2t^2+16t+1=0-2t^2+16t+1=0 put into the form  t^2+2at+a^2+b-2(t^2-8t+16-.5-16)=0-2(t-4)^2-.5-16=0-2(t-4)^2-16.5=0(t-4)^2=-16.52/2 t= +/-(?16.52+4)/2t=8.062t=-0.062In this case, total time is equal to 8.062 seconds when the ball lands (ignoring -0.062 because it isn’t relevant)Solving using the quadratic equation_y=-4.9t2+39.2t+2.45 we set the  equation equal to 0 and divide by negative like terms and to give us  whole and positive term value 0=2t^2-16t-1Using the quadratic formula we plug in the values of a, b and ct=(-(-16)^2 +/-?(-16)2-42(-1)22)/2*2Solving this equation we get the same answers: t=-0.062 and t=8.062I like completing the square and quadratic formula because they are  similar and allow to manipulate numbers that don’t necessarily factor  easilySource:Quadratic Applications: Projectile Motion (n.d). Retrieved from POST2:Over the weekend, I went on a canoe trip.  I spent the morning  traveling upstream and the afternoon traveling downstream.  Overall, I  spent 4 hours traveling 10 miles.  I know that the stream I was canoeing  flows at 3 MPH.  Now I need to find out at what speed I traveled. (distance upstream/speed – current) + (distance downstream/speed + current) = 4 hours(10 / x -3) + (10 / x+3) = 4 hours10 (x+3) + 10 (x-3) = 4 (x+3)(x-3)10x + 30 + 10x – 30 = 4(X^2 – 9)20X = 4x^2 – 9F(x) = 4x^2 – 20x – 9Quadratic Formula:X = (-b +/- ?(b^2-4ac)) / 2aX = (-(-20)+/- ?(-(-20)-4(4)(-20))) / ((2)(4))X = (20 +/- ?(400 + 144))) / 8X = (20 +/- ?(544))/8X = 5.41X = -0.41Therefore, since I cannot travel at a negative speed, I traveled at 5.41 MPH.Completing the square:4x^2 – 20x – 9            = 0(4/4)x^2 – (20/4)x – (9/4) = (0/4)X^2 -5x – (9/4) = 0X^2 – 5x = (9/4)X^2 – 5x + (25/4) = (9/4) + (25/4)(X – (5/2))^2 = (34/4)X – (5/2) = +/-?(17/2)X = (5/2) +/- ?(17/2)X = 5.41X = -0.41Overall, I preferred to use the quadratic formula.  I found that it  was easier for me to plug numbers into a formula versus what I saw as a  massive amount of work by completing the square.