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-it

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 \dots

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 .