anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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.
anonymous
  • anonymous
it then falls and its motion is described by another equation
anonymous
  • anonymous
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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@ArchiePhysics so wat is ur answer!
anonymous
  • anonymous
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
anonymous
  • anonymous
Do you have any questions?
anonymous
  • anonymous
|dw:1338391009065:dw|
anonymous
  • anonymous
@ArchiePhysics
anonymous
  • anonymous
Didn't understand, absolutely. What is that?
anonymous
  • anonymous
it s tan alpha
anonymous
  • anonymous
it is the answer
anonymous
  • anonymous
|dw:1338391169913:dw|
yash2651995
  • yash2651995
s tan theta
anonymous
  • anonymous
yes
yash2651995
  • yash2651995
wait.. full explanation is on the way..
anonymous
  • anonymous
it is the answer @yash2651995
anonymous
  • anonymous
@yash2651995 is that expl frm u??
yash2651995
  • yash2651995
uploading my solution
anonymous
  • anonymous
ok waiting
yash2651995
  • yash2651995
compressed picture. original is taking lots of time.. i'd still upload it later
1 Attachment
anonymous
  • anonymous
@yash2651995 bhai, thanks for the answer. Can u explain how?|dw:1338391988800:dw|
anonymous
  • anonymous
@yash2651995 thanzzz
anonymous
  • anonymous
Thanzz a billion @yash2651995 r u frm DElhi!!!!
yash2651995
  • yash2651995
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
yash2651995
  • yash2651995
@shakir neh yarr i'm from U.P.; a small suburb near Vns.
anonymous
  • anonymous
Got it yash bhai!!!!!!!! bahut bahut shukriya. UPwalon ke liye to IIT pakka hi hia na
yash2651995
  • yash2651995
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 ♪
anonymous
  • anonymous
hehheheheh. mera bhi wohi halat. bhagwan ki krupa se mil jaye ..........

Looking for something else?

Not the answer you are looking for? Search for more explanations.