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A Particle is projected upwards from ground with a speed u at an angle theta with the horizontal. At the same time another paricle is dropped from a height h if they collide at a horizontal distance s from the point of projection then h is?
 one year ago
 one year ago
A Particle is projected upwards from ground with a speed u at an angle theta with the horizontal. At the same time another paricle is dropped from a height h if they collide at a horizontal distance s from the point of projection then h is?
 one year ago
 one year ago

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ArchiePhysicsBest ResponseYou've already chosen the best response.0
At the moment of collision t = s/v1x , where v1x  initial horizontal component of velocity of the first object for the second object y(t) = h  gt^2/2 = h  gs^2/(2 u^2(cos theta)^2), but this is true only if the collision occurs before the first object rises its maximum height.
 one year ago

ArchiePhysicsBest ResponseYou've already chosen the best response.0
it then falls and its motion is described by another equation
 one year ago

ArchiePhysicsBest ResponseYou've already chosen the best response.0
Sorry I got little confused.That's all ok, we are not considering vertical motion of the first object, so the resultant equation is true for every moment of time. At the moment of collision t = s/v1x , where v1x  initial horizontal component of velocity of the first object v1x = u cos theta for the second object y(t) = h  gt^2/2 = h  gs^2/(2 u^2(cos theta)^2)
 one year ago

shakirBest ResponseYou've already chosen the best response.0
@ArchiePhysics so wat is ur answer!
 one year ago

ArchiePhysicsBest ResponseYou've already chosen the best response.0
The last equation is the answer y(t) = h  gs^2/(2 u^2(cos theta)^2) where y  is the height at which the second object is located h  is the initial height of it
 one year ago

ArchiePhysicsBest ResponseYou've already chosen the best response.0
Do you have any questions?
 one year ago

ArchiePhysicsBest ResponseYou've already chosen the best response.0
Didn't understand, absolutely. What is that?
 one year ago

yash2651995Best ResponseYou've already chosen the best response.2
wait.. full explanation is on the way..
 one year ago

shakirBest ResponseYou've already chosen the best response.0
it is the answer @yash2651995
 one year ago

shakirBest ResponseYou've already chosen the best response.0
@yash2651995 is that expl frm u??
 one year ago

yash2651995Best ResponseYou've already chosen the best response.2
uploading my solution
 one year ago

yash2651995Best ResponseYou've already chosen the best response.2
compressed picture. original is taking lots of time.. i'd still upload it later
 one year ago

AadarshBest ResponseYou've already chosen the best response.0
@yash2651995 bhai, thanks for the answer. Can u explain how?dw:1338391988800:dw
 one year ago

shakirBest ResponseYou've already chosen the best response.0
Thanzz a billion @yash2651995 r u frm DElhi!!!!
 one year ago

yash2651995Best ResponseYou've already chosen the best response.2
i'm replacing theta with O v cosO is horizontal component of initial velocity of particle 'a'.. which will remain unaltered as no force thus acceleration acts on it (the horizontal component) hence, by eq, s= ut +1/2 at^2 a=0 u=u cosO so s= u cosO t
 one year ago

yash2651995Best ResponseYou've already chosen the best response.2
@shakir neh yarr i'm from U.P.; a small suburb near Vns.
 one year ago

AadarshBest ResponseYou've already chosen the best response.0
Got it yash bhai!!!!!!!! bahut bahut shukriya. UPwalon ke liye to IIT pakka hi hia na
 one year ago

yash2651995Best ResponseYou've already chosen the best response.2
IIT ab toh teacher ke chamchon ke liye ho gaya hai :( ♪ ♫ apni toh naiyaa hai raam ke bharose ... apni bhi naiyya ko paar tu lagaai de ♪
 one year ago

AadarshBest ResponseYou've already chosen the best response.0
hehheheheh. mera bhi wohi halat. bhagwan ki krupa se mil jaye ..........
 one year ago
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