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 2 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?
 2 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?

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razor99
 2 years ago
Best ResponseYou've already chosen the best response.0wow is that a physics question

razor99
 2 years ago
Best ResponseYou've already chosen the best response.0this is the first time i ever saw a physics question lol cuz i study business stream

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2can u write a equation for the first stone distance travelled by it in t seconds

unkabogable
 2 years ago
Best ResponseYou've already chosen the best response.1i know that: \[V _{01}=0 ft/s\] \[V _{02}=248 ft/s\] \[g=32.2 ft/s^2\] \[t _{1}=t _{2}+2\]

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2@unkabogable can u?

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2So for the first stone distance travelled by it in t seconds is = ut + 1/2 at^2 = 16t^2

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2now find, for the first stone distance travelled by it in 2 seconds

unkabogable
 2 years ago
Best ResponseYou've already chosen the best response.1yeah...wait: so i have these eq. \[Y _{1}=1000Y _{2}\] \[Y _{2}=V _{02}t _{2}1/2(32.2)t _{2}^2\] then i'll substitute and i'll verify my answer later

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2Let 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
 2 years ago
Best ResponseYou've already chosen the best response.2Now,Now, after two seconds, height of the stone after t seonds =100064( ut+1/2 at^2) = 93664t  16t^2

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2then 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
 2 years ago
Best ResponseYou've already chosen the best response.2Now, when the the stones pass each other , 93664t  16t^2=248t  16t^2 936=312t t=3seconds

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2so the stone will pass each other at 3+2 seconds = 5 seconds

unkabogable
 2 years ago
Best ResponseYou've already chosen the best response.1yes..thanks again!! :)

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2No, can u find this where will the stones pass each other

unkabogable
 2 years ago
Best ResponseYou've already chosen the best response.1(248)(52)\[(248)(52) \frac{ (32.2(52)^2) }{ 2} = 600 ft\] ??

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2Yep........ 600 ft above the ground
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