A balloon is moving up from the ground in such a way that its acceleration is linearly decreasing with its height above the ground. It starts from the ground with acceleration 4 ml 52 and with zero initial velocity. Its acceleration becomes zero at a height 3 m. The speed of the balloon at a height 1.5 m is (A) 4 m/s (B) 8 m/s (C) 6 m/s (D) 3 m/s
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01. A balloon is moving up from the ground in such a way that its acceleration is linearly decreasing
with its height above the ground. It starts from the ground with acceleration 4 m/ s2 and with zero
initial velocity. Its acceleration becomes zero at a height 3 m. The speed of the balloon at a height
1.5 m is ……..
2 years ago
Answers : (2)
ACCELERATION IS LINEARLY DEPENDENT
SO,acc=ax+b
acc=4,b=4 for x=0.
acc=0 for a= -4/3 and x=3
and
acc =v*dv/dx
on integrating
\int_{0}^{v}vdv =\int_{0}^{3}(4-4x/3)dx
v=\sqrt{12}
2 years ago
if a is changing linearly(i.e.gradually with same rate),
=>da/dh=slope of the a vs h graph= (4-0)/(0-3) (as at h=0 =>a=4 and at h=3 =>a=0)
=>da/dh= −4/3
=>da=−(4/3)dh
on intergrating a=−(4/3)h +C
as a=4 => h=0
C=4
therefore,
a=−(4/3)h + 4
=>dv/dt=−(4/3)+4
=>(dv/dh)x(dh/dt)=−(4/3)h + 4 ….….…..we know dh/dt=v
=>(dv/dh)x(v)=−(4/3)h + 4
=>dv×v={−(4/3)h + 4}dh
on intergrating,
=>v2/2={−(2/3)h2 + 4h + c)
since v=0=>h=0
c=0
=>v2/2=−(2/3)h2 + 4h
when h=1.5
=>v2/2= 4×1.5 − (2/3)×1.52
=>v2/2=6−1.5=4.5
=>v2=9
=>v=3 m/s−1
the previous answer is incorrect because the user had intergrated x from 0 to 3 but it should have been 0 to 1.5 cos the v at 1.5 is asked in the question
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Answer:
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