SOLUTION: A small rock sits on the edge of a tall building. A strong wind blows the rock off the edge. The distance, in feet, between the rock and the ground tseconds after the rock leaves t

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Question 987005: A small rock sits on the edge of a tall building. A strong wind blows the rock off the edge. The distance, in feet, between the rock and the ground tseconds after the rock leaves the edge is given by d=−16t2−5t+470.
If the answer is not an integer, enter it as a decimal. Round to the nearest hundredth
How many seconds after the rock leaves the edge is it 446feet from the ground?How many seconds after the rock leaves the edge does it hit the ground?
Found 2 solutions by ankor@dixie-net.com, macston:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A small rock sits on the edge of a tall building.
A strong wind blows the rock off the edge.
The distance, in feet, between the rock and the ground t seconds after the rock leaves the edge is given by d = −16t^2 − 5t + 470.
If the answer is not an integer, enter it as a decimal. Round to the nearest hundredth
:
From the given formula we can surmise that the cliff is 470 above the ground below, which is the reference; d = 0
:
How many seconds after the rock leaves the edge is it 446feet from the ground?
−16t^2 − 5t + 470 = 446
−16t^2 − 5t + 470 - 446 = 0
−16t^2 − 5t + 24 = 0
Using the quadratic formula x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+, where
a = -16; b = -5; c = 24
I got a positive solution of s = 1.08 seconds
:
How many seconds after the rock leaves the edge does it hit the ground?
d = 0
−16t^2 − 5t + 470 = 0
Now a = -16; b = -5; c = 470
I got a positive solution of s = 5.27 seconds

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
-16t%5E2-5t%2B470=d
.
-16t%5E2-5t%2B470=446
-16t%5E2-5t%2B24=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation at%5E2%2Bbt%2Bc=0 (in our case -16t%5E2%2B-5t%2B24+=+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=%28-5%29%5E2-4%2A-16%2A24=1561.

Discriminant d=1561 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+1561+%29%29%2F2%5Ca.

t%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+1561+%29%29%2F2%5C-16+=+-1.39092164157115
t%5B2%5D+=+%28-%28-5%29-sqrt%28+1561+%29%29%2F2%5C-16+=+1.07842164157115

Quadratic expression -16t%5E2%2B-5t%2B24 can be factored:
-16t%5E2%2B-5t%2B24+=+-16%28t--1.39092164157115%29%2A%28t-1.07842164157115%29
Again, the answer is: -1.39092164157115, 1.07842164157115. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B-5%2Ax%2B24+%29

.
t=1.08 seconds
ANSWER:The rock is 446 feet from the ground after 1.08 seconds.
.
Rock hits ground when d-0
-16t%5E2-5t%2B470=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation at%5E2%2Bbt%2Bc=0 (in our case -16t%5E2%2B-5t%2B470+=+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=%28-5%29%5E2-4%2A-16%2A470=30105.

Discriminant d=30105 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+30105+%29%29%2F2%5Ca.

t%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+30105+%29%29%2F2%5C-16+=+-5.57837265284547
t%5B2%5D+=+%28-%28-5%29-sqrt%28+30105+%29%29%2F2%5C-16+=+5.26587265284547

Quadratic expression -16t%5E2%2B-5t%2B470 can be factored:
-16t%5E2%2B-5t%2B470+=+-16%28t--5.57837265284547%29%2A%28t-5.26587265284547%29
Again, the answer is: -5.57837265284547, 5.26587265284547. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B-5%2Ax%2B470+%29

.
t=5.27 seconds
ANSWER: The rock hits the ground after 5.27 seconds.