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 targ

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Question 34252: The distance d, in feet, a bomb falls in t seconds is given by d+=+%2816t%5E2%29%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 Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
d+=+%2816t%5E2%29%2F%281%2B.06t%29+
21000+=+%2816t%5E2%29%2F%281%2B.06t%29+
21000%281+%2B+.06t%29+=+%2816t%5E2%29+
21000+%2B+1260t+=+16t%5E2+
0+=+16t%5E2+-+1260t+-+21000+
0+=+4t%5E2+-+315t+-+5250+
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation at%5E2%2Bbt%2Bc=0 (in our case 4t%5E2%2B-315t%2B-5250+=+0) has the following solutons:

t%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-315%29%5E2-4%2A4%2A-5250=183225.

Discriminant d=183225 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--315%2B-sqrt%28+183225+%29%29%2F2%5Ca.

t%5B1%5D+=+%28-%28-315%29%2Bsqrt%28+183225+%29%29%2F2%5C4+=+92.8809868145612
t%5B2%5D+=+%28-%28-315%29-sqrt%28+183225+%29%29%2F2%5C4+=+-14.1309868145612

Quadratic expression 4t%5E2%2B-315t%2B-5250 can be factored:
4t%5E2%2B-315t%2B-5250+=+4%28t-92.8809868145612%29%2A%28t--14.1309868145612%29
Again, the answer is: 92.8809868145612, -14.1309868145612. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B-315%2Ax%2B-5250+%29

(C)