SOLUTION: if an object is propelled upward from a height of 64 feet at an initial velocity of 48 feet per second, then its height h after t seconds is given by the equation h=-16t^2+80t+48.

Algebra ->  Trigonometry-basics -> SOLUTION: if an object is propelled upward from a height of 64 feet at an initial velocity of 48 feet per second, then its height h after t seconds is given by the equation h=-16t^2+80t+48.       Log On


   

Question 675752: if an object is propelled upward from a height of 64 feet at an initial velocity of 48 feet per second, then its height h after t seconds is given by the equation h=-16t^2+80t+48. after how many seconds did the object hit the ground
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

h= -16t^2+80t+48 = 0   (object hitting the ground)

 -16t^2+80t+48 = 0

 16t^2-80t-48 = 0

 t^2-5t-3 = 0

t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

t+=+%285+%2B-+sqrt%28+37+%29%29%2F%282%29+ ||Positive value only for unit measure

0r

 -16(t-5/2)^2 + 100 + 48 = 0

 -16(t-5/2)^2 + 148 = 0

     t = 5/2 ± sqrt(148/16)

     t = 5/2 ± sqrt(37)/2