A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward followed again 5 steps forward and and 3 steps backward and so on. Each step is 1m long and requires 1sec. Determine how long the drunkard takes to fall in a pit 13m away from the stand.

hi..!! te answer is 49 sec

PANKAJ

IX-C

hi kreeti!!

i Tantuja.. i think it will take 37 sec for the drunkard to fall in the pit…

try again tantuja …!! you can do it

PANKAJ

IX-C

Hi Pankaj

Tantuja has arrived at the correct answer. This is more of a puzzle than being a mathematics/physics question.

Before putting simple algebra into it, just think about the last few steps of the drunkard…. Do you want him to jump over the pit and then return back 3 steps to fall into it? Won't he fall into it a long before that (while crossing it in forward motion!)… Please recheck your solution.

Wishes

Arpit

P. S. @ Kreeti and others, please try to give a catchy line in the Subject field and Summary of the thread. Otherwise, other students might not be attracted towards opening this link!

hi kreeti it will take it will take 42sec the drunkard takes to fall in a pit

A stone is thrown in avertically upward direction with a velocity of 5ms-1 . If the acceleration of the stone during its motion is 10 ms-2 in the downward direction , what will be the hight attained by the stone and how much time will it take to reach there ?

Anandita Dalal

ix-c

sol:-

v^{2}=u^{2}+2as

0 =25-2*10*s

0 =25-20s

20s = 25

s =25/20 = 5/4

s =1.25 m ans

PANKAJ

IX-C

Great my friend pankaj you have used the 3rd formula of motion

v^{2}- u^{2} = 2as

where we don't have t=time

ADITYA

IX C

good by first eqation:-

v=u+at

t=v-u/a

(0-5) ms-1/-10ms-2

=1/2s

t=0.5s

by shubhamchauhan

ix-c to anadita dalal