Question

a body falls from hieight H and travel 9h/25 in last second find height

Answer 1

Let's find the start velocity of the last second journey : time t = 1 sec, distance covered s = 9h/25 m

Using the equation : s = ut + 0.5gt^2

9h/25 = u(1) + 0.5*g*(1)^2

u = 9h/25 - g/2

This initial velocity u of the last second is the final velocity v of the journey prior to last sec.

For prior journey : v = 9h/25 - g/2

The start of this prior journey began with u = 0. Our aim is now to find how much distance s’ covered here -

v^2 = u^2 + 2gs’

(9h/25 - g/2)^2 = 0 + 2gs’

s' = (9h/25 - g/2)^2 / 2g

The total height of fall h is given by s' + 9h/25

h = s' + 9h/25

h= ((9h/25 + g/2)^2)/2g + 9h/25

This is a quadratic equation in h/g. Let h/g = y, then

(9y/25)^2 - 41y/25 + 1/4 = 0

Solving this quadratic equation gives two values

y1 = 25/2 and y2 = 25/162 (Both valid)

Resubstitute y = h/g

h/g = 25/2 and 25/162

h = 25g/2 and 25g/162

This two heights satisfy the condition that in the last second of their respective falls covers 9/25th fraction of their total heights.

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