contestada

The equation S = -16t2 + 34t + 184 models the height of a ball that is thrown upward from the roof of a 184 foot building and falls to the street below. In this equation S is the height in feet of the ball above the ground and t is the time in seconds the ball has traveled. According to this model, how many seconds did it take the ball to reach a height of 91 feet? (round to 1 decimal place)

Respuesta :

veronicaculs

Answer:

t=1.6

Step-by-step explanation:

  • If the equation [tex]S=-16t^2+34t+184[/tex] models the height of a ball above the ground, where t is time the ball travelled.
  • If the height of the ball above the ground is S=91 instead of 184, then the time the ball should take to get to the ground comes from the expression above: [tex]S=-16t^2+34t+91[/tex] (because now we want to know how much time does it takes to reach the ground if it is thrown from 91 foot, not 184).
  • Then, to know when the ball reaches the floor, we must equal the equation to zero [tex]-16t^2+34t+91=0[/tex] (because when the equation is zero, the height of the ball is zero, which means it is in the ground).
  • To obtain the value of t in the expression [tex]-16t^2+34t+91=0[/tex] , we can apply the well known formula [tex]t=\frac{-b(+-)\sqrt{b^2-4ac} }{2a}[/tex], where a is the coefficient that accompanies the quadratic term (in this case a=16), b is the coefficient that accompanies the linear term (b=34 in this case), and c is the constant coefficient (c=91).
  • Because time is always possitive, we only retain the possitive value for t that solves the equation:[tex]\frac{-34(+-)\sqrt{34^2-4\times(-16)\times93} }{2\times16}[/tex]. [tex]t=1.55\simeq1.6[/tex]