IIT JEE , AIEEE Forum - Physics , Mechanics

amulye

a body freely falling from a height h describes 7h/16 in d last second of its fall d height h is

45m giv a good explaination or better tell me d procedure

Decoder

look distance covered in Nth second is ...

S(n) = u + a/2 { 2n-1}..

here u is 0 as it is freely falling.
.a=g..

n is the nth second we r talikng abt..here nth second is the last second..
so n is nothing but the time taken to complete the fall...

this u can calculate by ..
H = 1/2 g n^2..

from here u get n..
substitute it in previous formula..

u get..
.. 7H/16 = g/2 { root[ 2H/g] - 1 }

solve for h now..

sriram.a

is the question right amulya

karthik2007

Method given by Decoder is correct. Follow it.

sriram.a

method is right but iam not geting answer

karansingh

u should be able to get the answer easily.

ayshwarya

na ive done d same procedure but its a vast calculation do u think dat in aieee dey wud giv dees type of calculations if do any1 have shortcut plz tell me

harsha_27

what is the question yaar????

what should we find in that????

ayshwarya

oyy v must find d h value

destinationIIT09

@AISHWARYA

well i feel there is a much easier way sloving dis in AIEEE,

if its obejective jus put the gievn heights n c wich suits

for eg:

h = 45m

hence t = (2h/g)^1/2 = 3 sec

distance covered in first 2 sec =1/2 g t^2

=5 * 4 = 20 m

now if distance covered in las second is 7h/16

CERTAINLY SUBSTITUION IS MUCH EASIER N ADOPTABLE IN AIEEE

ayshwarya

hey frnd but d substitution of 45m doesnt satisfy exactly may b approximately

amulye

common yaar rates surely assured

destinationIIT09

take g = 9.8,then it will

i hope so

plzzzz c

amulye

hey did u get it by doing such method

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