IIT JEE , AIEEE Forum - Physics , Mechanics

ssk317

From a tower of height y_o a particle is projected vertically up with a velocity v_oin free space(gravity free). The acc.of particle at an instant is given as a

k(v * A) vector, where v is instantaneous velocity and A is given as A= c/y (-k vector), where y is vertical coordinate of particle at that instant.(Take origin as foot of the tower)
For fig.

http://mypage.skrbl.com/ft5.html

Q.1 Speed of particle as a fn. of time will be:-
1. v_o + kt
2 v_o + y/t -1/2 at
3. v_o - y/t
4 v_o

Q.2 Maximum height attained by th particle......??

devvrat

Even i could not get it ...........

i got the 1st part

acc.. is perpendicular to velocity bcoz of the cross product..

so it will not do any work on it ...

aankurverma

v = V(x i +yj + zk)

A = c/y(-k)

a = K(v*A) = K c/y Vxj + K c /yVy(-i)

ay ax

ay = Kc/y Vx

-Kc/y root (vo)^2 - (vY)^2

ay = vy.dvy/dy

now solve d differential eqn n get d relation as

y - yo e^vo/kc

got it

hopes

Could you please explain the 4th step. how did you get ay

aankurverma

i cross k = j

elessar_iitkgp

The solution has been excellently provided by @aankurverma

Given,
v₀ = v₀ j
a = k(v x A)
A = (-c/y) k

Now,

a = (kc/y)v_x j - (kc/y)v_y i

v_y.dv_y/dy = (kc/y)v_x = (kc/y)

(v² - v_y²)
At maximum height, y_m, the y component of velocity is zero.

_v0

⁰ [v_y/

(v² - v_y²)]dv_y = kc _y0

^ym (1/y)dy
Integrating you should get the answer.

ssk317

Thanks ankur & elessar_iitkgp

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