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A stone dropped from the top of a tower is found to travel 5/9 of the height of the tower during the last second of its fall. the time of fall is?

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Let the height of tower be x

u = 0m/s

so let time to cover 5/9x of building be t sec

so time taken to cover 4x/9 of the building = t - 1 sec

5x/9= 5(2t - 1) ...........(i) g = 10

so

again s = ut +1/2gt²

4x/9 = 0 + 5(t - 1)² ..................(ii)

dividing i and ii

5x/9/4x/9 =5(2t - 1) /5(t - 1)²

⇒5t² +5 - 10t = 8t - 4

⇒5t² -18t +9 =0

solving t we get

t = 3 sec and 0.6 sec

u = 0m/s

so let time to cover 5/9x of building be t sec

so time taken to cover 4x/9 of the building = t - 1 sec

5x/9= 5(2t - 1) ...........(i) g = 10

so

again s = ut +1/2gt²

4x/9 = 0 + 5(t - 1)² ..................(ii)

dividing i and ii

5x/9/4x/9 =5(2t - 1) /5(t - 1)²

⇒5t² +5 - 10t = 8t - 4

⇒5t² -18t +9 =0

solving t we get

t = 3 sec and 0.6 sec

h=1/2 gt²

fall in t-1 time period = h- 5/9h= 4/9h

Thus, 4/9h = 1/2 g (t-1)²

dividing first equation by the second

t²/(t-1)² = 9/4

taking square root

t/(t-1)= 3/2

2t=3t-3

t=3 (answer)