Description
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?
Explanation & Answer
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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