The distance d in feet a bomb falls in t seconds is given by d=16t^2/1+0.06t . How many seconds are required for a bomb released at 15,000 feet to reach its target?

d = (16t^2)/(1+0.06t)

15000=(16t^2)/(1+0.06t)

15000(1+0.06t)=16t^2

15000+900t=16t^2

Move variables to one side:

16t^2-900t-15000=0

Simplify:

4(4t^2-225t-3750)=0

4t^2-225t-3750=0

This cannot be factored, so you must use the quadratic equation.

64(t-225/8)^2 / 110625 = 1

64(t-225/8)^2=110625

(t-225/8)^2=110625

Take the square root of both sides to cancel the ^2 term:

(t-225/8)=sqrt(110625/64)

sqrt(110625) simplifies to (25/8)*sqrt(177)

So the term becomes:

t-225/8=(25/8)*sqrt(177)

Now solve for t, and move 8 from denominator to numerator on the other side of the equation. This clears the 8 in (25/8):

t-225=25sqrt(177)

t=25sqrt(177)+225

Simplify!

t=sqrt(177)+9 or t=(25/8)sqrt(177)+9 if you don't want to simplify the term. This has only a positive solution, which solves to about

t=~69.7 seconds :)

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