Question

A stone is released from the top of a tower. If it falls half of the height of tower
in the last second of its journey, find the height of the tower and also calculate
the time of falling of stone.​

Answer 1

height of the tower  = 57.12  m   &  time of falling of stone.​  3.41  sec

Explanation:

Let Say tower height  = 2h  m

Distance (2h/2) = h  covered in last 1 sec

using S = ut + (1/2)at²

Here S = h  m

t = 1

a = g  = 9.8 m/s²

u = ?

h = u  + (1/2)9.8 (1)²

=> u  = h  - 4.9

from initial  drop to reach this point

this u will be now v  = h  - 4.9

Distance covered = h

initial velocity = 0  ( Dropped)

a = g = 9.8 m/s²

using

V² - U²  = 2aS

=> (h  - 4.9 )²  - 0  = 2(9.8)h

=> h²  - 9.8h  + 4.9² = 19.6h

=> h²  - 29.4 h  + 24.01 =  0

using(- b ± √b² - 4ac )/2a   for ax² + bx + c = 0

=> h = 28.56     as h can not less than 4.9

height of the tower  = 2 * 28.56  = 57.12  m

time of falling of stone

using S = ut + (1/2)at²  from initial drop

57.12 = 0 + (1/2)(9.8)t²

=>  t = 3.41  sec

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Answer 2

Explanation:

Given A stone is released from the top of a tower. If it falls half of the height of tower  in the last second of its journey, find the height of the tower and also calculate  the time of falling of stone.

• Given a stone is released which falls half the height in one second.
• Now we have a tower and when the stone is dropped the initial velocity is zero and also acceleration a = g.
• According to the question it falls half the tower in 1 sec. If we consider the height as h, then h/2 will be 1 sec.
• Now we have S = ut + 1/2 at^2
•         So h/2 = 0 + 1/2 g (1)^2
•       So h/2 = 1/2 g
•   Or h = g
• Or h = 9.8 m
• Now we need to find time for the entire tower.
• So total height will be  
• So h = ut + 1/2 gt^2
• So g = 0 + 1/2 x g x t^2
• So t = √2  secs
• So t = 1.414 secs
• So to cover total time it will take is 1.414 secs

Reference link will be

https://brainly.in/question/11484180