The building is 30 feet tall and
casts a shadow 8 feet wide in front of the store. if
kathy is 5 feet 6 inches tall, how far away from the
building can she stand and still be in the shade

hi there is think that I can help you with your problem!

h = -16t^ 2 + 20t + 6     the ball will hit the ground when h  = 0

0 = -16t^2  + 20t   + 6      factor

0  = ( -8t  -  2 )  ( 2t  - 3)     set each  factor to 0

-8t  -  2  =  0                                 and                  2t  - 3   = 0

-8t   = 2                                                                 2t   =   3

t  = -2/8  = - 1/4  sec   [reject]                                t  = 3/2 sec  = 1.5 sec

Max ht  will be reached at [ -b / 2a ]    sec    =  -20 / [ 2(-16)]  =  20/32  =  5/8 sec

Step-by-step explanation:  

-16t^2+20t+6 = 0

will give you the times that the ball is on the ground. The negative root accounts for the fact that the ball is 6 feet above ground at t=0.

I get t=3/2 and t=-1/4

-2(2t -3) (4t +1) = 0

2t -3 = 0

2t-3 + 3 = 0 + 3

2t = 3

t = 3/2 second        

I hope that I helped you and have a nice day.