SOLUTION: 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

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: 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       Log On


   

Question 34985: The distance d, in feet, a bomb falls in t seconds is given by d=16t%5E2%2F%281+%2B.06t%29
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
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The distance d, in feet, a bomb falls in t seconds is given by d=16t%5E2%2F%281+%2B.06t%29How 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.