Math help
- Thread starter Enjoi.
- Start date
tommyp1ckles
Active member
its really hard to read...
thelittlebiggs
Member
Sorry can't really see it
Stuck.in.ohio
Member
divide by zero
Enjoi.
Member
Ok it says
While it was flying 5 m above a lake, a kingfisher spotted a fish and immediately swooped down. It entered the water 2 seconds later. It caught the fish and then came back up.The kingfisher exited the water 2 seconds after entering it. It dropped the fish 1 second after coming out of the water.In the following cartesian plane a portion of a parabola represents the height of the king fisher above the surface of the lake in relation to the time elapsed from the moment it spotted the fish.
At what height above the surface of the lake was the kingfisher flying when it dropped the fish?
While it was flying 5 m above a lake, a kingfisher spotted a fish and immediately swooped down. It entered the water 2 seconds later. It caught the fish and then came back up.The kingfisher exited the water 2 seconds after entering it. It dropped the fish 1 second after coming out of the water.In the following cartesian plane a portion of a parabola represents the height of the king fisher above the surface of the lake in relation to the time elapsed from the moment it spotted the fish.
At what height above the surface of the lake was the kingfisher flying when it dropped the fish?
tommyp1ckles
Active member
from what i can tell it looks like a quadratic problem; if he uploaded a good image itd prob be really easy to solve
NaturalRotater
Member
You sure this is math and not physics?
EH_DUBS
Active member
you probably have to make an equation for the parabola.
y=a(x-3)^2-h...
fuck i dont even remember how to do this shit haha. all i know know is fucking trig identities.
it dosent say how deep it went in the water hey? i dont know how to figure out where to start the parabola.
y=a(x-3)^2-h...
fuck i dont even remember how to do this shit haha. all i know know is fucking trig identities.
it dosent say how deep it went in the water hey? i dont know how to figure out where to start the parabola.
Slinger1x1
Active member
butttt
hes flying at a constant speed and so it takes 2.5m/s to get into the water. when he exits, he is flying at a lesser speed and most likely accelerating again (i didnt look the graph at all yet), so therefore the 2.5m/s probably doesnt apply after he comes out of the water.
hes flying at a constant speed and so it takes 2.5m/s to get into the water. when he exits, he is flying at a lesser speed and most likely accelerating again (i didnt look the graph at all yet), so therefore the 2.5m/s probably doesnt apply after he comes out of the water.
Chickendodo
Active member
I think unless it's stated, the velocity would stay the same. This is really more of a Physics problem than a math problem...
Slinger1x1
Active member
how could velocity stay the same? if you hit water at 5m/s, in the water there is friction and therefor your velocity would decrease, and when you leave the water, you cannot be going as fast as you were when you hit unless you accelerate up to that speed again
wait it an african kingfisher? or european? that could help...
wait it an african kingfisher? or european? that could help...
Chickendodo
Active member
I lied (it's still a lot to do with physics, I swear). The answer is 1.875
You're given 3 points, (0,5), the initial point, (2,0), where the bird enters the water and (4,0), where the bird exits the water.
Then you do 3 equations of the form y = ax^2 + bx + c
first one comes out to c = 5
Then you just do simultaneous equations to find a and b
y = 4a + 2b + 5 (substitute 5 in for c)
y = 16a + 4b + 5
It figures out to a = 5/8 and b = -15/4
So the equation is then set to be y = (5/8)x^2 - (15/4)x + 5,
you plug in (5,?) and you get that the point where the bird drops the fish is (5, 15/8) or (5,1.875).
This was probably useless cause he already figured it out but oh well.
You're given 3 points, (0,5), the initial point, (2,0), where the bird enters the water and (4,0), where the bird exits the water.
Then you do 3 equations of the form y = ax^2 + bx + c
first one comes out to c = 5
Then you just do simultaneous equations to find a and b
y = 4a + 2b + 5 (substitute 5 in for c)
y = 16a + 4b + 5
It figures out to a = 5/8 and b = -15/4
So the equation is then set to be y = (5/8)x^2 - (15/4)x + 5,
you plug in (5,?) and you get that the point where the bird drops the fish is (5, 15/8) or (5,1.875).
This was probably useless cause he already figured it out but oh well.
Chickendodo
Active member
In math they disregard these variables all the time haha, I would know, I'm a math major. In lower level Physics they ignore them too, but once you start to do more advanced Physics they bring them into play. It's more of a thought experiment, where it has constant velocity.
Listen. In order to maintain air-speed velocity, a swallow needs to beat its wings forty-three times every second, right?
Listen. In order to maintain air-speed velocity, a swallow needs to beat its wings forty-three times every second, right?
