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

Algebra ->  Functions -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
Simply set the height equations equal to each other and solve for "t"


-16t%5E2+%2B+30=-8t+%2B+15 Set the two right sides equal to each other.


-16t%5E2%2B30%2B8t-15=0 Get all terms to the left side.


-16t%5E2%2B8t%2B15=0 Combine like terms.


Notice we have a quadratic in the form of At%5E2%2BBt%2BC where A=-16, B=8, and C=15


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


t+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


t+=+%28-%288%29+%2B-+sqrt%28+%288%29%5E2-4%28-16%29%2815%29+%29%29%2F%282%28-16%29%29 Plug in A=-16, B=8, and C=15


t+=+%28-8+%2B-+sqrt%28+64-4%28-16%29%2815%29+%29%29%2F%282%28-16%29%29 Square 8 to get 64.


t+=+%28-8+%2B-+sqrt%28+64--960+%29%29%2F%282%28-16%29%29 Multiply 4%28-16%29%2815%29 to get -960


t+=+%28-8+%2B-+sqrt%28+64%2B960+%29%29%2F%282%28-16%29%29 Rewrite sqrt%2864--960%29 as sqrt%2864%2B960%29


t+=+%28-8+%2B-+sqrt%28+1024+%29%29%2F%282%28-16%29%29 Add 64 to 960 to get 1024


t+=+%28-8+%2B-+sqrt%28+1024+%29%29%2F%28-32%29 Multiply 2 and -16 to get -32.


t+=+%28-8+%2B-+32%29%2F%28-32%29 Take the square root of 1024 to get 32.


t+=+%28-8+%2B+32%29%2F%28-32%29 or t+=+%28-8+-+32%29%2F%28-32%29 Break up the expression.


t+=+%2824%29%2F%28-32%29 or t+=++%28-40%29%2F%28-32%29 Combine like terms.


t+=+-3%2F4 or t+=+5%2F4 Simplify.


So possible the solutions are t+=+-3%2F4 or t+=+5%2F4

Since a negative time doesn't make sense, this means that the only solution is t+=+5%2F4 which is t=1.25


So the pelican will catch the crab in 1.25 seconds.


Notice how if we plug in t=1.25 into either equation, we get:

g%281.25%29=-8%281.25%29%2B15=-10%2B15=5 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) About Me  (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?
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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) About Me  (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
+graph%28+300%2C+200%2C+-4%2C+4%2C+-10%2C+30%2C+-16x%5E2%2B30%2C-8x%2B15%29+