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 During a "big air" competition, snowboarders launch themselves from a half-pipe, perform tricks in the air, and land back in the half-pipe. The height h​ (in feet) of a snowboarder above the bottom of the half-pipe can be modeled by h=-16t^2+24t+16.4, where t​ is the time (in seconds) after the snowboarder launches into the air. The snowboard lands 3.2​  feet lower than the height of the launch. How long is the snowboarder in the air? Round your answer to the nearest tenth of a second.

 Apr 30, 2021
 #1
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The initial  height   =  16.4  ft

 

So.....the  hright  at landing is    16.4 - 3.2 =  13.2  ft

 

We can solve this

 

13.2   =  -16t^2  +  24t  +  16.4       rearrange as

 

16t^2  - 24t  - 3.2  =    0

 

Using  the  quad  formula we  get

 

t  =   [ 24  +  sqrt ( 24^2  - 4 (15)(-3.2) )  ]  /  ( 2 * 16)      ≈ 1.6  seconds

 

 

cool cool cool

 Apr 30, 2021

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