Every year in delaware there is a contest where people create cannons and catapults designed to launch pumpkins as far in the air as possible. the equation y=10+95x-16x^2 can be used to represent the height, y, of a launched pumpkin, where x is the time in seconds that the pumpkin has been in the air. what is the maximum height that the pumpkin reaches? how many seconds have passed when the pumpkin hits the ground? (hint if the pumpkin hits the ground, its height is 0

Respuesta :

Answer:

max 151.02

after 6.04

Step-by-step explanation:

Using quadratic function concepts, it is found that:

  • The maximum height of the pumpkin is of 151 feet.
  • The pumpkin hits the ground after 6.04 seconds.

The height of the pumpkin after x seconds is modeled by:

[tex]y = -16x^2 + 95x + 10[/tex]

Which is a quadratic equation with coefficients [tex]a = -16, b = 95, c = 10[/tex].

The maximum height is the y-value of the vertex, given by:

[tex]y_{MAX} = -\frac{\Delta}{4a} = -\frac{b^2 - 4ac}{4a}[/tex]

Hence:

[tex]\Delta = b^2 - 4ac = (95)^2 - 4(-16)(10) = 9665[/tex]

[tex]y_{MAX} = -\frac{\Delta}{4a} = -\frac{9665}{4(-16)} = 151[/tex]

The maximum height of the pumpkin is of 151 feet.

It hits the ground when [tex]y = 0[/tex], hence:

[tex]x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-95 + \sqrt{9665}}{2(-16)} = -0.1[/tex]

[tex]x_2 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-95 - \sqrt{9665}}{2(-16)} = 6.04[/tex]

The pumpkin hits the ground after 6.04 seconds.

To learn more about quadratic functions, you can check https://brainly.com/question/24737967