## shakir 3 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?

1. ArchiePhysics

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.

2. ArchiePhysics

it then falls and its motion is described by another equation

3. ArchiePhysics

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)

4. shakir

@ArchiePhysics so wat is ur answer!

5. ArchiePhysics

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

6. ArchiePhysics

Do you have any questions?

7. shakir

|dw:1338391009065:dw|

8. shakir

@ArchiePhysics

9. ArchiePhysics

Didn't understand, absolutely. What is that?

10. shakir

it s tan alpha

11. shakir

12. shakir

|dw:1338391169913:dw|

13. yash2651995

s tan theta

14. shakir

yes

15. yash2651995

wait.. full explanation is on the way..

16. shakir

17. shakir

@yash2651995 is that expl frm u??

18. yash2651995

19. shakir

ok waiting

20. yash2651995

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

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

22. shakir

@yash2651995 thanzzz

23. shakir

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

24. 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

25. yash2651995

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

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

27. 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 ♪