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Question Number 44730 by Necxx last updated on 03/Oct/18

Commented by Necxx last updated on 03/Oct/18

$p l e a s e h e l p w i t h n o . 5$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Oct/18
	$s=ut+(1/2)gt^2 distance in t second distance covered in t−1 second s_(t−1) =u(t−1)+(1/2)g(t−1)^2 s−s_(t−1) =distance covered in last one second =u(t−t+1)+(1/2)g(t^2 −t^2 +2t−1) s_(last) =u+(1/2)g(2t−1) now s=h s_(last) =((9h)/(25)) (s_(last) /s)=(((9h)/(25))/h)=(((1/2)g(2t−1))/((1/2)gt^2 )) [ u=0 ] (9/(25))=((2t−1)/t^2 ) 9t^2 −50t+25=0 t=((50±(√(2500−900)))/(18)) t=(((50+40)/(18))), (((50−40)/(18))) t=5 and {(5/9)←not feasible} h=(1/2)×9.8×5^2 =122.5$
	$s = u t + \frac{1}{2} {g t}^{2} d i s t a n c e i n t s e c o n d$ $d i s t a n c e c o v e r e d i n t - 1 s e c o n d$ $s_{t - 1} = u (t - 1) + \frac{1}{2} g {(t - 1)}^{2}$ $s - s_{t - 1} = d i s t a n c e c o v e r e d i n l a s t o n e s e c o n d$ $= u (t - t + 1) + \frac{1}{2} g (t^{2} - t^{2} + 2 t - 1)$ $s_{l a s t} = u + \frac{1}{2} g (2 t - 1)$ $n o w s = h s_{l a s t} = \frac{9 h}{25}$ $\frac{s_{l a s t}}{s} = \frac{\frac{9 h}{25}}{h} = \frac{\frac{1}{2} g (2 t - 1)}{\frac{1}{2} {g t}^{2}} [u = 0]$ $\frac{9}{25} = \frac{2 t - 1}{t^{2}}$ $9 t^{2} - 50 t + 25 = 0$ $t = \frac{50 \pm \sqrt{2500 - 900}}{18}$ $t = (\frac{50 + 40}{18}), (\frac{50 - 40}{18})$ $t = 5 a n d {\frac{5}{9} \leftarrow n o t f e a s i b l e}$ $h = \frac{1}{2} \times 9.8 \times 5^{2} = 122.5$
	Commented by Necxx last updated on 04/Oct/18

	$y e a h . . . T h a n k y o u s o m u c h .$
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