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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?

Physics
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wow is that a physics question
obviously..yes :)
this is the first time i ever saw a physics question lol cuz i study business stream

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Other answers:

can u write a equation for the first stone distance travelled by it in t seconds
i know that: \[V _{01}=0 ft/s\] \[V _{02}=248 ft/s\] \[g=32.2 ft/s^2\] \[t _{1}=t _{2}+2\]
So for the first stone distance travelled by it in t seconds is = ut + 1/2 at^2 = 16t^2
right?
yes.
yes..
now find, for the first stone distance travelled by it in 2 seconds
can u?
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
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
Now,Now, after two seconds, height of the stone after t seonds =1000-64-( ut+1/2 at^2) = 936-64t - 16t^2
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
Now, when the the stones pass each other , 936-64t - 16t^2=248t - 16t^2 936=312t t=3seconds
so the stone will pass each other at 3+2 seconds = 5 seconds
got it?
yes..thanks again!! :)
WElcome
No, can u find this where will the stones pass each other
Now
(248)(5-2)\[(248)(5-2)- \frac{ (32.2(5-2)^2) }{ 2} = 600 ft\] ??
Yep........ 600 ft above the ground

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