deepak
Last Activity: 7 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=4
=>v2/2= 4×1.5 − (2/3)×1.52
=>v2/2=6−1.5=4.5
=>v2=9
=>v=3 m/s−1
p.s. dude pls dont do sums which are this difficult in tenth itself learn to enjoy life and after 11th or 12th do sums of these kinds, even for a twelth standard like me this sum was daunting XP
