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 2 years 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?
 2 years 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?

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ArchiePhysics
 2 years ago
Best ResponseYou've already chosen the best response.0At 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.

ArchiePhysics
 2 years ago
Best ResponseYou've already chosen the best response.0it then falls and its motion is described by another equation

ArchiePhysics
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry 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)

shakir
 2 years ago
Best ResponseYou've already chosen the best response.0@ArchiePhysics so wat is ur answer!

ArchiePhysics
 2 years ago
Best ResponseYou've already chosen the best response.0The 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

ArchiePhysics
 2 years ago
Best ResponseYou've already chosen the best response.0Do you have any questions?

ArchiePhysics
 2 years ago
Best ResponseYou've already chosen the best response.0Didn't understand, absolutely. What is that?

yash2651995
 2 years ago
Best ResponseYou've already chosen the best response.2wait.. full explanation is on the way..

shakir
 2 years ago
Best ResponseYou've already chosen the best response.0it is the answer @yash2651995

shakir
 2 years ago
Best ResponseYou've already chosen the best response.0@yash2651995 is that expl frm u??

yash2651995
 2 years ago
Best ResponseYou've already chosen the best response.2uploading my solution

yash2651995
 2 years ago
Best ResponseYou've already chosen the best response.2compressed picture. original is taking lots of time.. i'd still upload it later

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0@yash2651995 bhai, thanks for the answer. Can u explain how?dw:1338391988800:dw

shakir
 2 years ago
Best ResponseYou've already chosen the best response.0Thanzz a billion @yash2651995 r u frm DElhi!!!!

yash2651995
 2 years ago
Best ResponseYou've already chosen the best response.2i'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

yash2651995
 2 years ago
Best ResponseYou've already chosen the best response.2@shakir neh yarr i'm from U.P.; a small suburb near Vns.

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0Got it yash bhai!!!!!!!! bahut bahut shukriya. UPwalon ke liye to IIT pakka hi hia na

yash2651995
 2 years ago
Best ResponseYou've already chosen the best response.2IIT 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 ♪

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0hehheheheh. mera bhi wohi halat. bhagwan ki krupa se mil jaye ..........
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