A particle P of mass m, is projected vertically upward with a speed u from a point A, on horizontal ground. When P is at x above its initial position, its speed is v. The only forces acting on P is its weight and resistance mgkv^2 . where k is a positive constant. (a) Show that the greatest height reached is (1/(2gk)) ln(1 +ku^2 ). (b) show that the speed with which P returns to A is (u/(√(1+ ku^2 ))) .


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Question Number 96764 by Rio Michael last updated on 04/Jun/20

$A particle P of mass m, is projected vertically upward with$ $a speed u from a point A, on horizontal ground . When P is at x above$ $its initial position, its speed is v . The only forces acting on P is$ $its weight and resistance m g {k v}^{2} . where k is a positive constant .$ $(a) Show that the greatest height reached is \frac{1}{2 g k} \ln (1 + {k u}^{2}) .$ $(b) show that the speed with which P returns to A is \frac{u}{\sqrt{1 + {k u}^{2}}} .$
Commented by mr W last updated on 04/Jun/20

Commented by mr W last updated on 04/Jun/20

$n o f u r t h e r c o m m e n t . . .$
Commented by Rio Michael last updated on 05/Jun/20

$sorry sir, but on my phone it {doesn}^{'} t look that way .$ $its normal .$
Commented by mr W last updated on 06/Jun/20

$y o u h a v e a l a r g e s c r e e n s m a r t p h o n e .$ $t h e n {i t}^{'} s n o t p o s s i b l e f o r y o u t o$ $c o n t r o l t h e l i n e l e n g t h w h i l e t y p i n g .$
Commented by Rio Michael last updated on 08/Jun/20

$i^{'} ll try decreasing it . thanks$
	Answered by mr W last updated on 04/Jun/20

	$u p w a r d s :$ $m a = - m g - {m g k v}^{2}$ $a = \frac{d v}{d t} = - g (1 + {k v}^{2})$ $v \frac{d v}{d h} = - g (1 + {k v}^{2})$ $\frac{2 v k d v}{1 + {k v}^{2}} = - 2 g k d h$ $\frac{d (1 + {k v}^{2})}{1 + {k v}^{2}} = - 2 g k d h$ $\int_{u}^{0} \frac{d (1 + {k v}^{2})}{1 + {k v}^{2}} = - 2 g k \int_{0}^{h_{m a x}} d h$ ${[\ln (1 + {k v}^{2})]}_{u}^{0} = - 2 g k {[h]}_{0}^{h_{m a x}}$ $\Rightarrow h_{m a x} = \frac{1}{2 g k} \ln (1 + {k u}^{2})$ $d o w n w a r d s :$ $m a = m g - {m g k v}^{2}$ $a = \frac{d v}{d t} = g (1 - {k v}^{2})$ $- v \frac{d v}{d h} = g (1 - {k v}^{2})$ $\frac{- 2 v k d v}{1 - {k v}^{2}} = 2 g k d h$ $\frac{d (1 - {k v}^{2})}{1 - {k v}^{2}} = 2 g k d h$ $\int_{0}^{v_{A}} \frac{d (1 - {k v}^{2})}{1 - {k v}^{2}} = 2 g k \int_{h_{m a x}}^{0} d h$ ${[\ln (1 - {k v}^{2})]}_{0}^{v_{A}} = 2 g k {[h]}_{h_{m a x}}^{0}$ $\ln (1 - {k v}_{A}^{2}) = - 2 {g k h}_{m a x}$ $\ln (1 - {k v}_{A}^{2}) = - \ln (1 + {k u}^{2})$ $1 - {k v}_{A}^{2} = \frac{1}{1 + {k u}^{2}}$ $v_{A}^{2} = \frac{u^{2}}{1 + {k u}^{2}}$ $\Rightarrow v_{A} = \frac{u}{\sqrt{1 + {k u}^{2}}}$
	Commented by peter frank last updated on 04/Jun/20

	$thank you$
	Commented by Rio Michael last updated on 05/Jun/20

	$thank you sir .$
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