Question

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/s 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?​

Answer 1

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

HERE IS THE ANSWER

HOPE IT IS HELPFUL

Answer 2

I hope this is help full to you