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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?
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razor99
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wow is that a physics question
unkabogable
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obviously..yes :)
razor99
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this is the first time i ever saw a physics question lol cuz i study business stream
sauravshakya
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can u write a equation for the first stone distance travelled by it in t seconds
unkabogable
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i know that:
\[V _{01}=0 ft/s\]
\[V _{02}=248 ft/s\]
\[g=32.2 ft/s^2\]
\[t _{1}=t _{2}+2\]
sauravshakya
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@unkabogable can u?
sauravshakya
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So for the first stone distance travelled by it in t seconds is = ut + 1/2 at^2
= 16t^2
sauravshakya
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right?
unkabogable
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yes.
unkabogable
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yes..
sauravshakya
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now find, for the first stone distance travelled by it in 2 seconds
sauravshakya
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can u?
unkabogable
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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
sauravshakya
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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
sauravshakya
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Now,Now,
after two seconds,
height of the stone after t seonds =1000-64-( ut+1/2 at^2)
= 936-64t - 16t^2
sauravshakya
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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
sauravshakya
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Now, when the the stones pass each other ,
936-64t - 16t^2=248t - 16t^2
936=312t
t=3seconds
sauravshakya
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so the stone will pass each other at 3+2 seconds = 5 seconds
sauravshakya
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got it?
unkabogable
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yes..thanks again!! :)
sauravshakya
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WElcome
sauravshakya
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No, can u find this
where will the stones pass each other
sauravshakya
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Now
unkabogable
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(248)(5-2)\[(248)(5-2)- \frac{ (32.2(5-2)^2) }{ 2} = 600 ft\]
??
sauravshakya
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Yep........ 600 ft above the ground