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?

- anonymous

- schrodinger

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- 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

it then falls and its motion is described by another equation

- 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)

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- anonymous

@ArchiePhysics so wat is ur answer!

- 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

Do you have any questions?

- anonymous

|dw:1338391009065:dw|

- anonymous

- anonymous

Didn't understand, absolutely. What is that?

- anonymous

it s tan alpha

- anonymous

it is the answer

- anonymous

|dw:1338391169913:dw|

- yash2651995

s tan theta

- anonymous

yes

- yash2651995

wait.. full explanation is on the way..

- anonymous

it is the answer @yash2651995

- anonymous

@yash2651995 is that expl frm u??

- yash2651995

uploading my solution

- anonymous

ok waiting

- yash2651995

compressed picture. original is taking lots of time.. i'd still upload it later

##### 1 Attachment

- anonymous

@yash2651995 bhai, thanks for the answer. Can u explain how?|dw:1338391988800:dw|

- anonymous

@yash2651995 thanzzz

- anonymous

Thanzz a billion @yash2651995 r u frm DElhi!!!!

- 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

@shakir neh yarr i'm from U.P.; a small suburb near Vns.

- anonymous

Got it yash bhai!!!!!!!! bahut bahut shukriya. UPwalon ke liye to IIT pakka hi hia na

- 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

hehheheheh. mera bhi wohi halat. bhagwan ki krupa se mil jaye ..........

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