Grade 10Mechanics

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 ……..

MAHESWARA REDDY

10 Years agoGrade 10

2 Answers

vinay

10 Years ago

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

$\int_{0}^{v}vdv =\int_{0}^{3}(4-4x/3)dx$

$v=\sqrt{12}$

deepak

9 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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MAHESWARA REDDY

2 Answers

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deepak

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