IIT JEE , AIEEE Forum - Physics , Mechanics

sneha_91

from a elevated point A ,a stone is projected vertically upwards .when the stone reaches a height of h below A its velocity is double of what it was at a height h above the point A .find the maximum height reached by the stone above the point A

elessar_iitkgp

Let the point h below A be B an the point h above A be C
Taking upward direction positive,
v_C = +u (Say)
v_B = -2u
a = -g

y = -2h

Now applying the equation
v_B²=v_C²+ 2a(

y)
We get
u²= 4gh/3

Let velocity at A be u'
v_C²=v_A²+ 2a(

y')

v_A²= 10gh/3

Maximum height reached by the particle above A
H =v_A² /2g = 5h/3

sneha_91

WHEN U TAKE THE INITIAL VELOCITY AS X AND THE FINAL VELOCITY AS -2X THEN THE DISTANCE BETWEEN THEM IS -2HAND NOT -H

elessar_iitkgp

Ok will correct the solution

nick

what is the ans 2h/3???

sneha_91

maximum height upward so
v i.e final velocity is 0
v^2-u^2=2as
u^2=-40h/3
2a=-20
so s =-40h/3divided by -20
so distance traveled is 2h/3

sneha_91

yes nick u are right

nick

let the velocity h above be u

let it travel distance x more to reach max. height
therefore
x=u^2/2g.....

from that point to point h below the starting point
distance covered is 2h+x

final velocity is 2u
therefore,
2h+x=4u^2/2g

solving the two u get
x=2h/3
hence max. height is h+2h/3=5h/3

nick

but sneha ,
we have to count the distance 'h' as well!!!

elessar_iitkgp

Agree with Nick ... the answer is 5h/3

a_buddy4uin06

let "u" be the velocity with which it is projected
taki downward as positive dir...
v²-u²= -2gh(this is d equation wen d body travels upto h)
hence v²=u²- 2gh(this is d velocity at the point h above a)

now wen d body goes down to h, lets substitute the velocity has twice v
4(u²- 2gh)= u²+2gh
hence u²=10/3gh

now wen d body reaches d topmost position,
-u²=2as
substitutin fr u,
we get s as - 5h/3
thus dans..

cvramana

a_buddy4uin06 is correct.

zuma

i wann agive my reply'.

let the pt B be at a ht. of 'h' above A and the pt C be at a ht. of 'h' below A

let the velocity at B be ' v ',the velocity at C be ' 2v ',BC = 2h

(2v)^2 = v^2 + 2g(2h)

v^2 = 4gh/3

let D be the max.pt. above A,let AD be H

BD = H-h

to go from D to B, v^2 = 0 + 2g(H-h)

or 4gh/3 = 2g(H-h)

H = 2h/3 +h = 5h/3

rate me if u find it useful

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