A balloon rises from ground with an initial velocity of zero , and its acceleration decreases with height linearly . Its acceleration at ground is a0 and decreases to zero at height H0 then what will be its maximum velocity and velocity when it is at height H0/2 ?
Answers
Explanation:
When the stone is dropped from the balloon, its initial velocity is the same as the velocity of the balloon at that instant.
The upward acceleration of the balloon is a=2m/s
2
The balloon starts from rest, so u= 0 m/s
The balloon rises for 1 s before the stone is dropped, hence we have
v =u + at
v=0+=2 m/s in the vertically upw ard direction.
This is the initial velocity of the stone after being dropped.
The distance moved up by the balloon in 1 second is
v
2
=u
2
+2as
v
2
=2as
∴S=
2a
v
2
=
4
4
=1m
Hence, the stone falls by 1 m before hitting the ground
Now, the acceleration on the stone after being dropped is g=9.8m/s
2
Hence, for the stone,
we have
u=−2m/s;s=1m;a=9.8m/s
2
;t=1s
u is negative as it is in the upward direction
Hence, using third equation of motion,
we get time as
s=ut+
2
1
at
2
1=−2t+
2
1
×9.8t
2
4.9t
2
−2t−1=0
∴=t=
2×9.8
2
−
+
(−2)
2
−4×4.9×(−1)
=
2×9.8
2
−
+
4+19.6
=
9.8
2
−
+
23.6
Since, time cannot be negative, we have
t=
9.8
2+
23.6
=0.7s
Hence, the stone will strike the ground 0.7 s after being dropped from the balloon.