An object launched straight puwar from the ground level wiht an initional velocity of 32 feet per second. The formula H=32T-16T^2 gives it heigtht, h,in feet after t second. At what time does the object hit the ground?

2 secs after

Step-by-step explanation:

Given the height of the object modeled by the eqaution;

H(t) = 32T - 16T² where;

T is the time in seconds

The object hits the ground when h(t)= 0

Substitute H = 0 into the equation and get t as shown;

0 =  32T - 16T²

0-32T = -16T²

-32T = -16T²

32T = 16T²

32 = 16T

T = 32/16

T = 2secs

Hence the object hits the ground after 2 secs

step-by-step explanation:

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11