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Let the last second of the drop be expressed as t2 and the time before this as t1. So the total time, T, for the drop can be expressed as: T = (t1 +t2) or T = (t1 +1) (seconds).

During the last second, the object falls 75% of the height, h. This means that during the time before this: t1, it must have fallen 25% of the height.

S =(u x t) +0.5 g x t^2

Because the object is dropped from rest, the initial velocity u, will be zero.

thus S =0.5 g x t^2

(25 x h)/100 = 0.5 x g x t1^2 …..call this (1)

We can use S =(u x t) +0.5 g x t^2 again for the fall during the period t2. However, this time it already has some initial velocity (u) at the start of t2.The value of this velocity, v, will be v = g x t1. 

S =(u x t) +0.5 g x t^2

Substituting v=(g x t1) for u gives:-

(75 x h)/100= ((g x t1 x t2)+ (0.5 x g x t2^2))

But, since t2 =1, we end up with:-

(75 x h)/100= ((g x t1)+ (0.5 x g ))

(75 x h)/100= g x (t1 +0.5)…. Call this (2)

 To eliminate the unknown quantity,h,divide (1) by (2).

First, we do the left hand side (LHS) of the expressions:-

(25 x h)/100 divided by (75 x h)/100

(25 x h)/100 multiplied by 100/(75 x h)

(25 x h)/(75 x h)

LHS = 1/3

Now we do the RHS giving:-

=(0.5 x g x t1^2)/ g x (t1 +0.5)

=(0.5 x t1^2)/(t1 +0.5)

Now we equate both the LHS and RHS to complete the operation:-

1/3 = (t1^2/2) / (t1 +0.5)

2(t1 +0.5) = 3 x t1^2

2t1 +1 = 3t1^2

3t1^2 - 2t1 - 1 =0

Which is, of course a quadratic equation. 

We solve this quadratic using the ‘bog standard’ quadratic formula:-

(-b(+-) SQR ROOT of [b^2 - (4 x a x c)])/2a

and then substituting for: a = 3; b= -2 and c = -1 to give:-

(-(-2) (+-) SQR ROOT of [(-2^2) – (4 x 3 x (-1)])/ 2 x 3

(2 (+-) SQR ROOT of [4+12])/6

This gives us the solution as:- 

(2+ SQR ROOT of [16])/6 = 1 second.

t1 =1 second

The second solution being a invalid one.

So, t1 = 1 second.

Since  t2 was 1 second  the total time, T, of the drop was:-

T = t1 +1

T = 1 +1

T = 2 seconds.

From this we can calculate the total height of the tower by considering the ball falling from rest for 4 seconds. If we assume g = 10ms-2 then using S = 0.5 gT^2 we have:-

h = 0.5 x 10 x2^2

h = 20m.

The tower has a height of 20m.