A rocket is fired upward with an initial velocity v of 80 meters per second. The quadratic function S(t)=−5t2+80t can be used to find the height s of the​ rocket, in​ meters, at any time t in seconds. Find the height of the rocket 8 seconds after it takes off. During the course of its​ flight, after how many seconds will the rocket be at a height of 270 ​meters?

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  a) 320 m

  b) 4.838 seconds, or 11.162 seconds

Step-by-step explanation:

a) S(8) = -5(8^2) +80(8) = -320 +640 = 320

The height 8 seconds after takeoff will be 320 meters.

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b) S(t) = 270

  270 = -5t^2 +80t

  -54 = t^2 -16t . . . . . . divide by -5

  10 = t^2 -16t +64 . . . . add 64

  ±√10 = t -8 . . . . . . . . . . take the square root

  t = 8 ±√10 ≈ {4.838, 11.162}

The rocket will be at a height of 270 meters at 4.838 and 11.162 seconds after takeoff.

6 (3) - 4 + 32

hope this

its not b

step-by-step explanation: