Question

A ballon is moving up from the ground in such a way that it's acceleration is linearly decreasing with its height above the ground . It starts from the ground with acceleration
$4m ^{2}$
and with zero initial velocity . Its acceleration becomes 0 at height 3m. The speed of balloon at height 1.5m is :

(a) 4m/s (b) 8m/s (c) 6m/s (d) 3m/s

​

Answer 1

Answer:

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

=>dvxv={−(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= 4x1.5 − (2/3)x1.52

=>v2/2=6−1.5=4.5

=>v2=9

=>v=3 m/s−1

Explanation:

please follow me and like me and vote me