I remember doing A level maths about a thousand years ago, but can't for the life of me remember how to do these. Can anyone help solve this? I need working too, or just an explanation how to do it so I can solve it myself.
I hit a baseball directly over 2nd base into shallow centre field (largely irrelevant baseball reference)
1. How far does the ball fly before hitting the ground?
2. How long was the ball in the air?
The height of the ball as a function of distance relative is:
h = -0.05d^2 + 2.8d + 1
The height of the ball as a function of time is:
h = -6.3t^2 + 31.4t + 1
Note:
- ^ means to the power of.
- You need to use the Quadratic Formula to solve this.
Wtf is a quadratic formula?
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Cool, that helped a lot. I plugged all the numbers in the equation and came up with:
Distance function
h=1.99, h=23.77
Time function
h=1.98, h=0
Now to answer the actual question. How do these relate to each other to get the total distance before hitting the ground and time it was in the air? My brain is approaching melt down.
Distance function
h=1.99, h=23.77
Time function
h=1.98, h=0
Now to answer the actual question. How do these relate to each other to get the total distance before hitting the ground and time it was in the air? My brain is approaching melt down.
http://plus.maths.org/issue30/features/ ... -gifd.html
The bit next to the picture of an ill car looks like it could do the job
The bit next to the picture of an ill car looks like it could do the job
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bomberesque
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Friz
quadratic equations give you 2 solutions for each parameter when H=0, in your case the answers are;
Distance function
d=1.99, d=23.77
Time function
t=1.98, t=0
This is because the ball is at height = 0 twice in its arc. At the end of its arc and (confusingly) just before you throw it (the formula gives the whole curve, arcing back from your throwing arm to the ground behind you, as if the ball came from there at some small negative value of t.
I think you got the numbers wrong out of teh quad equation though, I believe they are
when H=0, t = -0.3 or 5.02 and d = -0.35 or 56.35
The negative figures are the ones corresponding to the solution of teh ball having come, not from your arm, but from behind you along the same trajectory and are as such a nonsense result. The ones you want are the positive ones, thus the ball flies for 5.02 seconds and goes 56.35, as it's baseball, it's probably feet
to confirm, plug the figures back into your original formula and you should get H=0 in both cases, if not then I made an error and you should run the numbers through the quad equation again
can you tell I'm bored?
btw, I've just realised, this is to do with trejectories of nondirect fire, isbn't it? this has more to do with cannons than baseball, I'd warrant
quadratic equations give you 2 solutions for each parameter when H=0, in your case the answers are;
Distance function
d=1.99, d=23.77
Time function
t=1.98, t=0
This is because the ball is at height = 0 twice in its arc. At the end of its arc and (confusingly) just before you throw it (the formula gives the whole curve, arcing back from your throwing arm to the ground behind you, as if the ball came from there at some small negative value of t.
I think you got the numbers wrong out of teh quad equation though, I believe they are
when H=0, t = -0.3 or 5.02 and d = -0.35 or 56.35
The negative figures are the ones corresponding to the solution of teh ball having come, not from your arm, but from behind you along the same trajectory and are as such a nonsense result. The ones you want are the positive ones, thus the ball flies for 5.02 seconds and goes 56.35, as it's baseball, it's probably feet
to confirm, plug the figures back into your original formula and you should get H=0 in both cases, if not then I made an error and you should run the numbers through the quad equation again
can you tell I'm bored?
btw, I've just realised, this is to do with trejectories of nondirect fire, isbn't it? this has more to do with cannons than baseball, I'd warrant
Okay well my solution (may have been solved, didn't bother to scroll down):
We have two equations:
-0.05d^2 + 2.8d + 1 = 0
-6.3t^2 + 31.4t + 1 = 0
Since the coefficient, in both of the equations, of the quadratic member is negative, we're going to get parabolas which have a maximum.
First, let's see if the equations have real solutions (though I doubt that balls fly for negative time or distance in baseball)
Dd = b^2-4ac = 2.8^2 + 4*0.05 = 7.84 + 0.2 = 8.04 > 0 -> two real solutions
Dt = 31.4^2 + 4*6.3 = 985.96 + 25.2 = 1011.16 > 0 -> two real solutions
Okay, let's use the Quadratic Formula on both of the equations
(note: sqrt means square root)
d1 = (-b+sqrt(b^2-4ac))/2a = (-2.8 + sqrt(8.04)) / -0.1 = (-2.8 + 2.835) / -0.1 = 0.035 / -0.1 = -0.35
d2 = (-b+sqrt(b^2-4ac))/2a = (-2.8 - sqrt(8.04)) / -0.1 = (-2.8 - 2.835) / -0.1 = -5.635 / -0.1 = 56.35
t1 = blah blah = -0.3
t2 = blah blah = 5.02
At this point I bothered to scroll around. Note that I might suck at counting though and may have messed something up. My results don't really seem to match any of the above. I tend to fuck these up all the time. Fix'd.
Now, this shows that when the ball hits the ground after 5.02 seconds, it should land at a 56.35 lolwuts distance from you.
We have two equations:
-0.05d^2 + 2.8d + 1 = 0
-6.3t^2 + 31.4t + 1 = 0
Since the coefficient, in both of the equations, of the quadratic member is negative, we're going to get parabolas which have a maximum.
First, let's see if the equations have real solutions (though I doubt that balls fly for negative time or distance in baseball)
Dd = b^2-4ac = 2.8^2 + 4*0.05 = 7.84 + 0.2 = 8.04 > 0 -> two real solutions
Dt = 31.4^2 + 4*6.3 = 985.96 + 25.2 = 1011.16 > 0 -> two real solutions
Okay, let's use the Quadratic Formula on both of the equations
(note: sqrt means square root)
d1 = (-b+sqrt(b^2-4ac))/2a = (-2.8 + sqrt(8.04)) / -0.1 = (-2.8 + 2.835) / -0.1 = 0.035 / -0.1 = -0.35
d2 = (-b+sqrt(b^2-4ac))/2a = (-2.8 - sqrt(8.04)) / -0.1 = (-2.8 - 2.835) / -0.1 = -5.635 / -0.1 = 56.35
t1 = blah blah = -0.3
t2 = blah blah = 5.02
At this point I bothered to scroll around. Note that I might suck at counting though and may have messed something up. My results don't really seem to match any of the above. I tend to fuck these up all the time. Fix'd.
Now, this shows that when the ball hits the ground after 5.02 seconds, it should land at a 56.35 lolwuts distance from you.
Last edited by Baliame on December 18th, 2007, 15:34, edited 2 times in total.
Coolio. I got to the same answer as you guys, but I confess after brainsploding all over my desk after trying to do it manually, I confess I cheated and Googled a nifty Quadratic Equation Solver. Love t'internets
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Roman Totale
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