two bodies a and b of masses 100g and 200g respectively are dropped near earth's surface let the acceleration of A and B be a1 and a2 respectivly
YourFriend1:
are u asking for their acceleration
more like an mcq
options are
a) a1=a2 b) a1<a2 c)a1>a2 d)a1=/=a2
option a)
Answers
Answered by
45
Are u asking for their acceleration?
If yes then their acceleration will be the same(I.e 9.8m/s^2) because acceleration due to gravity doesn't depend on mass of the object
If yes then their acceleration will be the same(I.e 9.8m/s^2) because acceleration due to gravity doesn't depend on mass of the object
more like an mcq
options are
a) a1=a2 b) a1<a2 c)a1>a2 d)a1=/=a2
options are
a) a1=a2 b) a1<a2 c)a1>a2 d)a1=/=a2
btw thankyou (brother or Sister)
option a)a1=a2 if u found my answer useful mark it as the BRAINLIEST
option for brainliest not here (brother or sister)
Brother
ok
Answered by
24
if the question has asked to find the time when they reach the bottom then it will be same for both A and B .
as we know H = 0t + 1/2 gt^2 or t = (2H/g)^2
● Only the impulse of heavier body will be more .
Their acceleratn has to be Same .
if Suppose they are a1 and a2 . Then it may be due to air friction and also air is different in the places where A and B are dropped.
● Therefore in that case consider net acceleration to be a1 and a2 .
a1 = g - (accln due to air friction in atmosphere1)
a2 = g - (accln due to air friction in atmosphere 2)
as we know H = 0t + 1/2 gt^2 or t = (2H/g)^2
● Only the impulse of heavier body will be more .
Their acceleratn has to be Same .
if Suppose they are a1 and a2 . Then it may be due to air friction and also air is different in the places where A and B are dropped.
● Therefore in that case consider net acceleration to be a1 and a2 .
a1 = g - (accln due to air friction in atmosphere1)
a2 = g - (accln due to air friction in atmosphere 2)
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