Quiz - Maths problem
13 March 2005, 12:24 AM
#31
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Originally Posted by Regulator
Well, ms is a measure of speed and not time so that makes little sense to me
13 March 2005, 12:25 AM
#32
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Well given both bullets start at the same height, and weigh the same, it can't be too hard to work out how long bullet two takes to hit the ground if you are given the time for bullet one. The rest of the info given is superfluous 
13 March 2005, 12:26 AM
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Originally Posted by carl
No, m/s or ms^-1 is a unit of velocity (metres per second). ms is a unit of time: milliseconds.
13 March 2005, 12:27 AM
#34
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Originally Posted by custard puddle
how? if a number is multiplied by itself no times, it would be 0, same as anything multiplied by 0 is 0.
It is clear that x^n = (x^(n+1))/x
Plug in n=0 and you get x^0 = (x^1)/x
i.e. x^0 = x/x = 1
So for all x, x^0 = 1
13 March 2005, 12:27 AM
#35
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Originally Posted by Hanslow
Well given both bullets start at the same height, and weigh the same, it can't be too hard to work out how long bullet two takes to hit the ground if you are given the time for bullet one. The rest of the info given is superfluous
13 March 2005, 12:28 AM
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Originally Posted by carl
Think of it as the following:
It is clear that x^n = (x^(n+1))/x
Plug in n=0 and you get x^0 = (x^1)/x
i.e. x^0 = x/x = 1
So for all x, x^0 = 1
It is clear that x^n = (x^(n+1))/x
Plug in n=0 and you get x^0 = (x^1)/x
i.e. x^0 = x/x = 1
So for all x, x^0 = 1
13 March 2005, 12:30 AM
#37
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Originally Posted by custard puddle
ooooooooooooooooooooooooooo to confuse people! well it worked on me! but if one bullet was travelling horizontally and the other was dropped, then surely the gravity pull and the speed of the 1st bullet would be different? oh i dont know, I'm just confusing myself even more now lol
Last edited by Hanslow; 13 March 2005 at 12:32 AM.
13 March 2005, 12:34 AM
#38
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Originally Posted by Hanslow
Nope, you are only interested in the time for each bullet to fall, so you can eliminate travel in the horizontal direction (assuming uniform air resistance, etc.) so you are only bothered about travel in the vertical plane. Given that you are told bullet one falls in x secs, it follows that a bullet of the same weight dropped from the same height will also take x secs to fall. The accelerative force due to gravity, i.e. 9.8055 m/s^2, is constant (for simple things like this). Any ballistics problem can be separated out into calculations in a horizontal and vertical plane.
what does the ^ mean?
13 March 2005, 12:42 AM
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It means squared. Couldn't be arsed to try and work out if I could do a little 2 in superscript 
I remember doing (and loving) ballistics problems at school. You'd often have to separate each problem into the vertical and horizontal. Then applying the equations of motion would allow you to tie in the scant information you were often given, such as maybe only a vertical height but no horizontal length of travel. You'd then calculate the time for the projectile to fall in the vertical plane, which you could then plug into the horizontal aspect and work out the other missing bits.....fond memories
I remember doing (and loving
) ballistics problems at school. You'd often have to separate each problem into the vertical and horizontal. Then applying the equations of motion would allow you to tie in the scant information you were often given, such as maybe only a vertical height but no horizontal length of travel. You'd then calculate the time for the projectile to fall in the vertical plane, which you could then plug into the horizontal aspect and work out the other missing bits.....fond memories
13 March 2005, 12:44 AM
#42
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Originally Posted by Hanslow
It means squared. Couldn't be arsed to try and work out if I could do a little 2 in superscript
I remember doing (and loving) ballistics problems at school. You'd often have to separate each problem into the vertical and horizontal. Then applying the equations of motion would allow you to tie in the scant information you were often given, such as maybe only a vertical height but no horizontal length of travel. You'd then calculate the time for the projectile to fall in the vertical plane, which you could then plug into the horizontal aspect and work out the other missing bits.....fond memories
I remember doing (and loving
) ballistics problems at school. You'd often have to separate each problem into the vertical and horizontal. Then applying the equations of motion would allow you to tie in the scant information you were often given, such as maybe only a vertical height but no horizontal length of travel. You'd then calculate the time for the projectile to fall in the vertical plane, which you could then plug into the horizontal aspect and work out the other missing bits.....fond memories
13 March 2005, 12:34 PM
#48
Originally Posted by Regulator
Well, ms is a measure of speed and not time so that makes little sense to me, but
s = ut+ 1/2 at²
1.75 = 0 + 1/2 X 10 X t²
time = 5.91 seconds
(and yes i do A level mechanics)
s = ut+ 1/2 at²
1.75 = 0 + 1/2 X 10 X t²
time = 5.91 seconds
(and yes i do A level mechanics)
Yours says that it takes a bullet 5.91 seconds to fall 1.75m? Nah
Surely, when you get to
1.75 = 0 + 1/2 X 10 X t², the next line is:
1.75= 5t^
1.75/5= t^
0.35 =t^, so t is the square root of 0.35, or 0.5916079s then 592ms to the nearest ms?
13 March 2005, 12:45 PM
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Originally Posted by alcazar
But your teacher never told you to LOOK at the answer and see if it MAKES SENSE then?
Yours says that it takes a bullet 5.91 seconds to fall 1.75m? Nah
Surely, when you get to
1.75 = 0 + 1/2 X 10 X t², the next line is:
1.75= 5t^
1.75/5= t^
0.35 =t^, so t is the square root of 0.35, or 0.5916079s then 592ms to the nearest ms?
Yours says that it takes a bullet 5.91 seconds to fall 1.75m? Nah
Surely, when you get to
1.75 = 0 + 1/2 X 10 X t², the next line is:
1.75= 5t^
1.75/5= t^
0.35 =t^, so t is the square root of 0.35, or 0.5916079s then 592ms to the nearest ms?
13 March 2005, 01:59 PM
#52
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Originally Posted by matt85
Correct.
13 March 2005, 02:09 PM
#53
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Originally Posted by custard puddle
how though? if something is to the power of something, it means it is multiplied by itself that many times, and if something is multiplied by itself NO times, how is it 1? was just wondering thats all...
or look at it like this:-
1. x^3 = x * x * x
2. x^2 = x * x
3. x^1 = x
Notice at each step the answer equals the previous answer divided by x
so to go from x^1 to x^0 we need to divide the previous answer by x again, which gives us x/x = 1
Therefore, any number to the power zero equals 1!
13 March 2005, 11:41 PM
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Originally Posted by ajm
See the proof that Carl posted on page 2!
or look at it like this:-
1. x^3 = x * x * x
2. x^2 = x * x
3. x^1 = x
Notice at each step the answer equals the previous answer divided by x
so to go from x^1 to x^0 we need to divide the previous answer by x again, which gives us x/x = 1
Therefore, any number to the power zero equals 1!
or look at it like this:-
1. x^3 = x * x * x
2. x^2 = x * x
3. x^1 = x
Notice at each step the answer equals the previous answer divided by x
so to go from x^1 to x^0 we need to divide the previous answer by x again, which gives us x/x = 1
Therefore, any number to the power zero equals 1!
