Question

A body is freely dropped from top of building. If it covers 36% of its total distance in its last second then determine the height of building!

(Need a detailed, step by step explained solution)

Answer 1

Let the height of the building be h
Then distance travelled in last second = $h \times \frac{36}{100}$
Time required to reach the ground = $\sqrt{ \frac{2h}{g} }$ seconds
We know distance travelled in $n^{th}$ second = $S_{n} = u+ \frac{1}{2} a(2n-1)$
Distance travelled in last second = distance travelled in $(\sqrt{ \frac{2h}{g} })^{th}$ second = $S_{\sqrt{ \frac{2h}{g} }} = 0+ \frac{1}{2} g(2\times(\sqrt{ \frac{2h}{g} })-1)$ = $\frac{36h}{100}$

Hence we get
[tex] \frac{1}{2}g(2 \times \sqrt{ \frac{2h}{g} }-1 ) = \frac{36h}{g}\\ Lets take g = 10 m/s^2 \\ Therefore \\ 5(2 \times \sqrt{ \frac{h}{5} }-1 ) = \frac{36h}{100} = \frac{9h}{25} \\ (2 \times \sqrt{ \frac{h}{5} }-1 ) = \frac{9h}{125}\\ (2 \times \sqrt{ \frac{h}{5} }) = \frac{9h}{125}+1 = \frac{9h+125}{125}\\ \sqrt{ \frac{h}{5} } = \frac{9h+125}{250}\\ \\ Squaring \\ \\ \frac{h}{5} = \frac{81h^2+2250h+15625}{62500} \\ h = \frac{81h^2+2250h+15625}{12500} \\ \\ 12500h = 81h^2+2250h+15625 \\ [/tex]
[tex]81h^2-10250h+15625 = 0 \\ solving \\ h = 125 m [/tex]

Answer 2

Full distance = -gt²/2 

Distance In last second  : 

Velocity in (t -1)sec

 U = -g(t -1) V = -,gt 

g²t² = g²(t -1)² -2gS 

g{ 2t -1} = 2S

S = g(2t -1)/2

 1/2 gt² × 36/100= g(2t -1)/2

 36t²/100 = (2t -1)

 36t² -200t + 100 = 0

9t² - 50t + 25 = 0

t = 50 +40/18,  t = 5 

Hence,  H = 1/2 × 10 × 25 = 125 m