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Mathematics: Post your doubts here!

GorgeousEyes

GorgeousEyes

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a particle p is projected vertically upwards from o with velocity 40 m/s one second later another particle is projected from O with the same vertical velocity .after what time and at what heoght will the two particles collide?
 
iKhaled

iKhaled

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GorgeousEyes said:
a particle p is projected vertically upwards from o with velocity 40 m/s one second later another particle is projected from O with the same vertical velocity .after what time and at what heoght will the two particles collide?
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from where did u get this question ?
 
GorgeousEyes

GorgeousEyes

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iKhaled said:
i am tryin to do it :S:S r u sure they have the same velocity and r they travellin in the same line? can u pls show me the question itself
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Unfortunately , there's no drawing for it :S . it's ok if you can't do it , i might ask my teacher about it and tell you later incha'allah .
 
VelaneDeBeaute

VelaneDeBeaute

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XPFMember said:
AsSalamoAlaikum Wr Wb


tan (A + B) = (tan A + tan B)/(1 - (tan A)(tan B))

from the info given, rearrange, and you get: x = tan A ; y = tan B

just substitute, and you get x + y / 1 - xy
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WS!
Sorry I still don't get it. :(
If arctan A = x, that means tan x = A, right? :confused:
 
iKhaled

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GorgeousEyes said:
Unfortunately , there's no drawing for it :S . it's ok if you can't do it , i might ask my teacher about it and tell you later incha'allah .
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yeah please do tell me about it cuz its driving me crazy.

howa ante masrya 3ysha f masr ?
 
whitecorp

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bamteck said:
Yeah I agree with you !
But the answer isn't right I guess :(
Have a look in the marking scheme !
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_ms_42.pdf
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Sorry my bad, actually the answer scheme is correct. A continues to move upwards even as B remain on the floor.
Since the string has slackened, tension in string now =0 N and the velocity of A decreases from 1.6 m/s to 0m/s under the influence of gravity.
Let additional distance moved by A=s2
Then using v^2 =u^2 +2a(s2), we have 0^2 =1.6^2 -2(10)(s2) =====>s2=0.128m
Total distance travelled by A = 0.128+0.32 =0.448m (shown)

Hope this clears things up .Peace.
 
XPFMember

XPFMember

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VelaneDeBeaute said:
WS!
Sorry I still don't get it. :(
If arctan A = x, that means tan x = A, right? :confused:
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Oops..! Yes you're right, that was a slip in my answer.

Well, then tell me what's your final answer.
I am 110% sure, if the question is as you mentioned, the answer in the book is wrong. Have ways to check that. :p
 
bamteck

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whitecorp said:
Sorry my bad, actually the answer scheme is correct. A continues to move upwards even as B remain on the floor.
Since the string has slackened, tension in string now =0 N and the velocity of A decreases from 1.6 m/s to 0m/s under the influence of gravity.
Let additional distance moved by A=s2
Then using v^2 =u^2 +2a(s2), we have 0^2 =1.6^2 -2(10)(s2) =====>s2=0.128m
Total distance travelled by A = 0.128+0.32 =0.448m (shown)

Hope this clears things up .Peace.
Click to expand...

No worries :) Thanks !
 
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