Meghraj Shah

• Sep 5th, 2005

 

576 cm

bhargavi

• Jan 22nd, 2006

 

since height is proportional to square of the time,

h1/h2=(t1*t1)/(t2*t2)

now let the height to be found be h2

64/h2=(2*2)/(6*6)

h2=576cm

MAHAVEER PATIL

• Feb 3rd, 2006

 

height of fall in 6 second will be 192 cms

anshul

• Mar 2nd, 2006

 

I am not able to understand the question itself....what is trying to say.If height is proportional to sq of time the how can object fall 64 in 6 sec ..shoud not it be 8 sec is sq root of 64... Can anyone explan this question in a simple and clear manner.

rony banerjee

• Mar 7th, 2006

 

The falling height is proportional to the squere of the time.

Now,the falling height is 64cm at 2sec

so,  the proportional constant is=64/(2*2)=16;

so, at 6sec the object fall maximum (16*6*6)cm=576cm;

Now, the object may be situated at any where .

if it is>576 only that time the object falling 576cm within 6sec .Otherwise if it is situated<576 then it fall only that height at 6sec. 

dheepti

• Sep 15th, 2006

 

576cm

coolgy1234

 Profile Answers by coolgy1234 

• Oct 28th, 2010

 

Ht - t^2

64-((root(2))^2

so

64 cm-2 sec

i.e 1sec=32cm., so in 6sec=6*32=192 Which is required answer - hp it is correct

Cheers.

sami_sharma

 Profile Answers by sami_sharma 

• Dec 17th, 2010

 

  ans:576 cm
  64=k4
so k=16
h=16*36

prititripathi

 Profile Answers by prititripathi 

• Dec 18th, 2010

 

given h=t^2

so by formula

h1/h2=(t1^2/t2^2)

64/h2=4/36

h2=576

Anshuman

• Oct 1st, 2020

 

F proportional to T ^2
so, f=kT^2 (where T is const.)
64=k2^2
k=64/4=16
then in 6sec falling distance (fd)=16*6^2
16*36=576cm(ans.)

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