## anonymous 4 years ago a stone is dropped from a captive balloon at an elevation of 1000ft,. two sec later another stone is projected vertically upward from the ground with a velocity of 248ft per sec. if g is 32ft per sec squared when and where will the stones pass each other?

1. razor99

wow is that a physics question

2. anonymous

obviously..yes :)

3. razor99

this is the first time i ever saw a physics question lol cuz i study business stream

4. anonymous

can u write a equation for the first stone distance travelled by it in t seconds

5. anonymous

i know that: $V _{01}=0 ft/s$ $V _{02}=248 ft/s$ $g=32.2 ft/s^2$ $t _{1}=t _{2}+2$

6. anonymous

@unkabogable can u?

7. anonymous

So for the first stone distance travelled by it in t seconds is = ut + 1/2 at^2 = 16t^2

8. anonymous

right?

9. anonymous

yes.

10. anonymous

yes..

11. anonymous

now find, for the first stone distance travelled by it in 2 seconds

12. anonymous

can u?

13. anonymous

yeah...wait: so i have these eq. $Y _{1}=1000-Y _{2}$ $Y _{2}=V _{02}t _{2}-1/2(32.2)t _{2}^2$ then i'll substitute and i'll verify my answer later

14. anonymous

Let me do it then, for the first stone distance travelled by it in 2 seconds=16*2^2 = 64ft now velocity of it after two seconds = (o^2 + 2*32 *64)^1/2 = 64ft/s

15. anonymous

Now,Now, after two seconds, height of the stone after t seonds =1000-64-( ut+1/2 at^2) = 936-64t - 16t^2

16. anonymous

then for the next stone: height of the stone after t seconds = ut + 1/2 at^2 = 248t + 1/2 (-32) t^2 =248t - 16t^2

17. anonymous

Now, when the the stones pass each other , 936-64t - 16t^2=248t - 16t^2 936=312t t=3seconds

18. anonymous

so the stone will pass each other at 3+2 seconds = 5 seconds

19. anonymous

got it?

20. anonymous

yes..thanks again!! :)

21. anonymous

WElcome

22. anonymous

No, can u find this where will the stones pass each other

23. anonymous

Now

24. anonymous

(248)(5-2)$(248)(5-2)- \frac{ (32.2(5-2)^2) }{ 2} = 600 ft$ ??

25. anonymous

Yep........ 600 ft above the ground