IIT JEE , AIEEE Forum - Physics , Mechanics - freefall-1

sneha_91

a stone dropped from the top of a tower describes during the last second of its fall 5/9th of the whole height .find the height of the tower?

edison

Let the height of the tower = h

Total time taken by stone to fall on the ground = t

so, in t seconds

h = (1/2) g t^{2 ..........}(1)

Now in (t-1) seconds distance travelled is given by

h - 5h/9 = (1/2) g (t -1)²

or 4h / 9 = (1/2) g (t -1)² ........(2)

Now taking the ratio of (1) and (2)

9/4 = t²/ (t -1)²

or 9 (t² + 1 - 2t) - 4 t² = 0

or 5 t² - 18t + 9 = 0

or 5 t² - 15 t - 3t + 9 = 0

or 5t (t - 3) - 3 (t - 3) = 0

or t = 3 and t = 3/5

so take t = 3

thus, from (1)

h = (1/2) g t²

or h = 0.5 x 10 x 3 x 3

or = 45m is the height of the tower

elessar_iitkgp

Let the total time of motion be T and the height of the tower be H
Then distance fallen till T-1 time = H - (5/9)H = (4/9)H
So,
(4/9)H = (1/2)g(T-1)²----------(1)
H = (1/2)gT² ---------(2)

Substituting (2) in (1)
5 T² - 18T + 9 = 0
T = 3 and T = 3/5

But the stone falls for atleast one second as per the problem
So T = 3
Hence H = 45 m

sneha_91

why do u subtract 5/9th h from h because distance traveled in t-1 sec will be 5/9th h

omkarnet

h=45 m

just refer gupta and gupta

elessar_iitkgp

5h/9 is the distance travelled in the last one second and not upto T-1th sec

So distance travelled till T-1 th sec = h - 5h/9

elessar_iitkgp

Does this help??

sneha_91

yes .thankyou as i didnot notice during the last second.
can u suggest a good physics book .please give some tips for my iit preparations

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