A balloon ascends with a constant acceleration of g/8.At the end of 30 sec a body is released from balloon. Find time taken by body to reach ground?
at the end of 30 sec speed of baloon is v=30g/8(using v=u+at formula)
so speed of thhe body released is also 30g/8in the upward direction
the body will fall with normal acc due to gravity i.e g
vel at the end of 30 sec is 30g/8
and the height attend is 1125/2 m
now apply formula s=ut+1/2at^2....
the answer is 19.21 sec
when stone is released from balloon its height is h=1/2at^2 = 1125/2
and vel 30g/8
time taken by stone to reach the ground..
apply this formula and get the result
at the end of 30 sec the velocity of balloon is 300/8 and distaance travelled is 9000/16
now use the eqution-----
9000/16= -300/8t +1/2 g t^2
simplified equation-----2t^2 -15t -225=0 solving this we get 15 sec as ans
Q) A balloon ascends with a constant acceleration of g/8.At the end of 30 sec a body is released from balloon. Find time taken by body to reach ground?
Sol) t = 30s, u = 0, a = g/8
thus using S = ut + 1/2 a*t2
S = 0.5 * g/8 * 900 = 551.25 m
Moreover v = u + at = g/8 * 30 = 15g/4 m/s is the velocity of the body in upward direction when dropped from the baloon after 30s
Thus, time taken by body to fall the ground is given by
t1 + t2
where t1 = time taken by the body to come back to the point from where it was released
t2 = time taken by body to reach the gound after coming back to the point from where it was released
For Finding t1:
use v - u = at
or 15g/4 - (-15g/4) = gt1
or t1 = 30/4 = 7.5 s
For Finding t2:
Use S = ut + 0.5 a t2
or 551.25 = (15g / 4)*t2 + 0.5 * g * t22
Find t2 and obtain total time t1 + t2
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