SOLUTION: Please HELP!! A pelican flying in the air over water drops a crab from a height of 30 feet. The distance the crab is from the water as it falls can be represented by the functi

Question 198727: Please HELP!!
A pelican flying in the air over water drops a crab from a height of 30 feet. The distance the crab is from the water as it falls can be represented by the function h(t)= -16t^2 + 30, where t is time in seconds. To catch the crab as it falls, a gull flies along a path represented by the function
g(t)= -8t + 15. Can the gull catch the crab before the crab hits the water?
Thanks!
Found 3 solutions by jim_thompson5910, Alan3354, ankor@dixie-net.com:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Simply set the height equations equal to each other and solve for "t"

Set the two right sides equal to each other.

Get all terms to the left side.

Combine like terms.

Notice we have a quadratic in the form of where , , and

Let's use the quadratic formula to solve for "t":

Start with the quadratic formula

Plug in , , and

Square to get .

Multiply to get

Rewrite as

Add to to get

Multiply and to get .

Take the square root of to get .

or Break up the expression.

or Combine like terms.

or Simplify.

So possible the solutions are or

Since a negative time doesn't make sense, this means that the only solution is which is

So the pelican will catch the crab in 1.25 seconds.

Notice how if we plug in into either equation, we get:

which means that the pelican will meet with the crab 5 feet in the air.

So the gull will catch the crab before the crab hits the water.
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
A pelican flying in the air over water drops a crab from a height of 30 feet. The distance the crab is from the water as it falls can be represented by the function h(t)= -16t^2 + 30, where t is time in seconds. To catch the crab as it falls, a gull flies along a path represented by the function g(t)= -8t + 15. Can the gull catch the crab before the crab hits the water?
------------------
There should be additional specs for this problem, but a simplification is to see if the 2 functions intersect.
No horizontal info is given, so solve to see if and when
-16t^2 + 30 = -8t + 15
-8t^2 + 15 = 0
8t^2 = 15
t^2 = 15/8
The paths intersect (vertically) at t = sqrt(30)/4 seconds
t = ~ 1.369 seconds from the release of the crab
at a height of 0 feet, at the surface of the water.
The crab and the gull meet the water at the same time, not before.
Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
A pelican flying in the air over water drops a crab from a height of 30 feet.
The distance the crab is from the water as it falls can be represented by the
function h(t)= -16t^2 + 30, where t is time in seconds.
To catch the crab as it falls, a gull flies along a path represented by the function g(t)= -8t + 15.
Can the gull catch the crab before the crab hits the water?
:
When the gull catches the crab, they will be at the same height, so we can say:
Gull ht = crab ht
-8t + 15 = -16t^2 + 30
16t^2 - 8t + 15 - 30 = 0
16t^2 - 8t - 15 = 0
Factors to
(4t - 5)(4t + 3) = 0
Positive solution
4t = 5
t = 1.25 sec they will be at the same height (5 ft)
:
A graph would would show this

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