SOLUTION: A type of aquatic bird known as a cormorant is flying over open water and accidentally drops a small fish from a height of 30 feet. The distance the fish is from the water as it fa

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Question 1042255: A type of aquatic bird known as a cormorant is flying over open water and accidentally drops a small fish from a height of 30 feet. The distance the fish is from the water as it falls can be described by the function h(t)=-16t^2+30 where t= time in seconds. A seagull spies the falling fish and flies along a straight line to intercept the fish. His line of flight represented by the linear equation g(t)=-8^t+15
PART 1 After how many seconds does the seagull catch the falling fish?
PART 2 After the seagull catches the fish he changes his travel path to k(t)=-2t+12.The cormorant chases the seagull in an attempt to try to win back the fish,descending along the path -t^2+15.How many feet above the water will the cormorant catch up to the seagull?
Found 2 solutions by ikleyn, Boreal:
Answer by ikleyn(52747) About Me  (Show Source):
You can put this solution on YOUR website!
.
A type of aquatic bird known as a cormorant is flying over open water and accidentally drops a small fish from a height of 30 feet.
The distance the fish is from the water as it falls can be described by the function h(t)=-16t^2+30 where t= time in seconds.
A seagull spies the falling fish and flies along a straight line to intercept the fish. His line of flight represented by the
linear equation g(t)=-8^t+15
PART 1 After how many seconds does the seagull catch the falling fish?
PART 2 After the seagull catches the fish he changes his travel path to k(t)=-2t+12.The cormorant chases the seagull in an attempt
to try to win back the fish,descending along the path -t^2+15.How many feet above the water will the cormorant catch up to the seagull?
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Sorry, the equation g(t) = -8%5Et%2B15 is not linear. 

If it is g(t)=-8t+15, then the solution for Part 1) is

-16t%5E2%2B30 = -8t%2B15,

16t%5E2+-+8t+%2B+15 = 0,

t = %288+%2B-+sqrt%2864%2B4%2A16%2A15%29%29%2F32 = %288+%2B-+32%29%2F32.

Only positive root is of interest: t = 5%2F4 seconds.

Answer to Part 1).  t = 5%2F4 seconds.





Plot h(t) = -16t%5E2%2B30 (red) and g(t) = -8t%2B15 (green)


Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
h(t) is the fish pathway
g(t) is the seagull, and I am editing that to -8t+15, since -8^t is not linear.
The two have the same value when the seagull catches the fish. So, set them equal to each other.
-16t^2+30=-8t+15
0=16t^2-8t-15
t^2-8t-240
(t-20)(t+12)
(t-20/16)(t+12/16)
t-(5/4))(t+3/4)
(4t-5)(4t+3)=0
t=5/4 sec, only positive root.
We need the distance above the water. It is- 8(5/4)+15=5 feet. Or, -16(25/16)+30=5 feet.
After catching the fish, new function -2t+12=-t^2+15
t^2-2t-3=0
(t-3)(t+1)=0
t=3,-1
use only the positive root.
Substituting 3 into either equation, 6 feet is the result.
The seagull started 5 feet above the water, so the cormorant catches up at 11 feet.