Question 34985
The distance d, in feet, a bomb falls in t seconds is given by  {{{d=16t^2/(1 +.06t)}}}How many seconds are required for a bomb released at  21,000 feet to reach its target? (If necessary, round your answer to two decimal places.) 

A)  1167.12 seconds
B)   185.76 seconds
C)  92.88 seconds
D)   2972.19 seconds 

d=16t^2/(1 +.06t)
21000(1+0.06t)=16t^2
21000+1260t-16t^2=0
4t^2-315t-5250=0
Graphing this I find the zero at 92.88 seconds.
Cheers,
Stan H.