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.
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) (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 ||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