## AravindG 5 years ago two bodies fall freely from different heights and arrive at earth in same time.the time of fall of first body is 2 s and other is 1 s .At what height was first body when second began to fall??

1. AravindG

2.A particle slides from rest from topmost point of a vertical circle of radius r along a smooth chord .Show that the time of descent is same irrespective of chord chosen and find it in terms of g and r

2. anonymous

$\mathsf{h = u_y\cdot t + \frac{1}{2} \cdot (g) \cdot t^2}$ I am assuming the bodies had Zero initial velocity. $\mathsf{ h = \frac{1}{2} \cdot g \cdot t^2 }$I think you can take it from here.

3. AravindG

hey but bodies cover different heights

4. anonymous

For different time. Just put in the value of time for the bodies separately.

5. AravindG

cnfused

6. anonymous

'bodies' sounds a like harsh word for physics. Okay, so we have two Objects A and B. Height or the distance traversed by the Objects is given by the following equation. $\mathsf{h_{i} = \frac{1}{2}\cdot g \cdot t_{i}^2}$ For the object A, let's assume time taken to land is 2s and for object B 1s. $\mathsf{h_{A} = \frac{1}{2}\cdot g \cdot t_{A}^2 = \frac{g}{2} \cdot 2^2}$ $\mathsf{h_{B} = \frac{1}{2}\cdot g \cdot t_{B}^2} = \frac{g}{2} \cdot 1$

7. anonymous

Note. The $$\mathsf{h_{i} = \frac{1}{2}\cdot g \cdot t_{i}^2}$$ equation is only valid if the initial velocity of the object 'i' is Zero. 

8. anonymous

Hey aravind, do you have a figure (diagram) for the second question?

9. AravindG

no

10. AravindG

pls answr both the qns

11. anonymous

Oh, I thought 1s and 2s is the time taken by each body to land and we have to determine their initial height accordingly. Thanks!

12. anonymous

Ishaan. The bodies are released at different times. Otherwise, there is no way they could reach the ground at the same time. (At least under Newton's Laws, we won't get into special relativity.) Let's express the height as a function of time and velocity. $h_1(t,v(t'))=v(t')⋅t+{1 \over 2}⋅g⋅t^2$ The height of the first object at the time the second object is released can be expressed as $h_1(2,0)−h_1(1,v(1))$

13. anonymous

I think I got a sign wrong here.

14. anonymous

But what about the second question? I don't understand a thing.

15. anonymous

I don't understand it either. A figure would be helpful.

16. JamesJ

I think it's asking this: along any such chord as in this diagram, the time for the object to reach the the point where the circle and chord meet is equal. Indians ask the damnest questions in Physics. |dw:1327424415886:dw|

17. JamesJ

Hint: this polar form an the equation of a circle or radius R will be helpful: $r = 2A\cos \theta$

18. anonymous

Oh I see, thanks James. I think I can do it now.

19. JamesJ

**correction: "of radius A" [not R!]

20. AravindG

k hw to do?

21. JamesJ

Repost this as a new question. It's worth everyone seeing this one.

22. AravindG

james k can u help me in permutations its urgent??

23. AravindG

pls

24. AravindG