SOLUTION: A model rocket is launched with an initial velocity of 160 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by h = −16t2 + 160t. How many secon

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A model rocket is launched with an initial velocity of 160 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by h = −16t2 + 160t. How many secon      Log On


   

Question 1139880: A model rocket is launched with an initial velocity of 160 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 160t.
How many seconds after launch will the rocket be 380 ft above the ground? Round to the nearest hundredth of a second.
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
when −16t2 + 160t = 0

Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.
It will be at the given height when  h(t) = 380,  or, equivalently,


    -16t%5E2+%2B+160t = 380


    16Tt%5E2+-160t+%2B+380 = 0.


Solve this quadratic equation


    t%5B1%2C2%5D = %28160+%2B-+sqrt%28160%5E2+-+4%2A16%2A380%29%29%2F%2816%2A2%29 = %28160+%2B-+35.78%29%2F32.


There are two solution, both make sense, t%5B1%5D = %28160+-+35.78%29%2F32 = 3.88 seconds in the increasing part of the parabola,

and  t%5B2%5D = %28160+%2B+35.78%29%2F32 = 6.12 seconds in the decreasing branch of the parabola.

Solved.