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Question Number 78496 by TawaTawa last updated on 18/Jan/20

Commented by mathmax by abdo last updated on 18/Jan/20

$m i s s t a w a a r e a t h e a t c h e r o r e n g i n e e r . . .$
Commented by TawaTawa last updated on 18/Jan/20

$Engineer sir .$
Commented by mathmax by abdo last updated on 18/Jan/20

$o h g o o d j o b m i s s t a w a . . w i c h y o u g o u d l u c k . . .$
Commented by TawaTawa last updated on 19/Jan/20

$Thank you sir, i appreciate .$
	Answered by mr W last updated on 18/Jan/20

	$m a = - m g - K m v$ $a = - g - K v$ $v \frac{d v}{d h} = - g - K v$ $\frac{v d v}{g + K v} = - d h$ $\int_{u}^{0} \frac{v d v}{g + K v} = - \int_{0}^{H} d h$ ${[\frac{v}{K} - \frac{g \ln (g + K v)}{K^{2}}]}_{u}^{0} = - H$ $- \frac{g \ln g}{K^{2}} - \frac{u}{K} + \frac{g \ln (g + K u)}{K^{2}} = - H$ $\Rightarrow H = \frac{u}{K} - \frac{g}{K^{2}} \ln (1 + \frac{K u}{g}) = z_{m a x}$
	Commented by peter frank last updated on 18/Jan/20

	$p l e a s e m r w h e l p 77889$
	Commented by TawaTawa last updated on 18/Jan/20

	$God bless you sir$
	Commented by mr W last updated on 18/Jan/20

	$s o r r y, i {c a n}^{'} t h e l p .$
	Commented by peter frank last updated on 18/Jan/20

	$w h y s i r o r t h e q u e s t i o n$ $i s n o t c l e a r ?$
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