IIT JEE , AIEEE Forum - Physics , Mechanics - plz solve (calculus based laws of motion qstn)

savvej

A particle is shot down vertically with a speed "u".the
resistance is kv^2/g
per unit mass in gravitational unit .v is instantaneous velocity .If
g=kC^2,prove that ,when the particle moves 'x' ,then v^2=C^2 - (C^2 -
u^2)e^-2kx.if the initial velocity is C/2 ,show that the time it takes to
obtain 3c/4 speed is ln(7/3)/2Ck.

shubham_sachdeva

i didn't get the qn. completely..what is C?? & also k?

well what i hv got above is dis:-

ma = mg - m.k.v^2/g

so

a = g - k.v^2/g

also, g = k.C^2

so a becomes,

a = k.C^2 - (v/C)^2 ----------1

now just applying eqn. of motion v^2 = u^2 + 2as

we get

v^2 = u^2 + 2.x.(k.C^2 - {v/C})^2

now just rearranginf we get

v^2 = u^2 + 2kxC^2

1 + 2x

C^2

i don't get what is v .. v is instantaneous velocty u write but at what instant....?

spideyunlimited

@ shubham
resistance is kv^2/g per unit mass in gravitational unit .v is instantaneous velocity ( in the given drag resistance eqn)

raltz

i am getting

v²= c⁴(1+2kx) / c²+2x

raltz

i am getting

v²= c⁴(1+2kx) / c²+2x

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