SOLUTION: A football is kicked straight up into the air. Its height above the ground is approximated by the relation y=25x - 5x^2 ,where y is the height of the ball in meters and x is the t

Algebra ->  Graphs -> SOLUTION: A football is kicked straight up into the air. Its height above the ground is approximated by the relation y=25x - 5x^2 ,where y is the height of the ball in meters and x is the t      Log On


   

Question 1195453: A football is kicked straight up into the air. Its height above the ground is approximated by the relation y=25x - 5x^2
,where y is the height of the ball in meters and x is the time in seconds.
There are two times when the football is 27.5 m in the air, on the way up and on the way down.
Joe realizes that using the equation
y= 27.5
, he can determine the two times the ball is 27.5
meters in the air. Algebraically or graphically determine these two times.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
y=25x+-+5x%5E2
if y=27.5
27.5=25x+-+5x%5E2
5x%5E2-+25x%2B27.5=0+........simplify, divide by 5
x%5E2-+5x%2B5.5=0+...........use quadratic formula

x+=+%28-%28-5%29+%2B-+sqrt%28+%28-5%29%5E2-4%2A1%2A5.5+%29%29%2F%282%2A1%29+
x+=+%285+%2B-+sqrt%28+25-22+%29%29%2F2+
x+=+%285+%2B-+sqrt%28+3+%29%29%2F2+

x+=+%285+%2B-+1.732%29%2F2+

solutions:
x+=+%285+%2B+1.732%29%2F2+ =>x+=+3.37
or
x+=+%285+-+1.732%29%2F2+=>x+=+1.63+

these two times are: +1.63+seconds and +3.37 seconds