A balloon rises from ground with an
initial velocity of zero, and its
acceleration decreases with height
linearly. Its acceleration at ground is ao
and decreases to zero at height Ho then find the maximum velocity of the balloon.Also find it's velocity when it is at height Ho/2.
Answers
Answer:
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