Small-steel-balls-falls-from-rest-through-the-opening-at-A-at-the-steady-rate-of-n-balls-per-second-Find-the-vertical-separation-h-of-two-consecutive-balls-when-the-lower-one-has-dropped-d-meters-

Question Number 15102 by Tinkutara last updated on 07/Jun/17

Small steel balls falls from rest through

the opening at A, at the steady rate of

n balls per second . Find the vertical

separation h of two consecutive balls

when the lower one has dropped d

meters .

Commented by Tinkutara last updated on 07/Jun/17

Answered by mrW1 last updated on 07/Jun/17

ball 1 :

d = \frac{1}{2} {gt}_{1}^{2}

t_{1} = \sqrt{\frac{2 d}{g}}

ball 2 :

d_{2} = \frac{1}{2} g {(t_{1} - \frac{1}{n})}^{2}

h = d - d_{2} = \frac{1}{2} g [t_{1}^{2} - {(t_{1} - \frac{1}{n})}^{2}]

= \frac{1}{2} g ({2 t}_{1} - \frac{1}{n}) (\frac{1}{n})

= \frac{g}{{2 n}^{2}} (2 n \sqrt{\frac{2 d}{g}} - 1)

Commented by Tinkutara last updated on 07/Jun/17

Thanks Sir!

Answered by ajfour last updated on 07/Jun/17

Δ t = \frac{1}{n}; l e t s_{i} b e t h e h e i g h t

t h r o u g h w h i c h i^{t h} b a l l h a s f a l l e n

a t a c e r t a i n i n s t a n t o f t i m e . L e t

a l s o t h a t t h e d u r a t i o n o f i t s f a l l

t i l l t b a t t i m e i n s t a n t i s t .

s_{i} = \frac{1}{2} {g t}^{2} = d

\Rightarrow 2 t = \sqrt{\frac{8 d}{g}} \dots . (i)

t h e n s_{i + 1} = \frac{1}{2} g {(t - Δ t)}^{2}

Δ h = s_{i} - s_{i + 1}

= \frac{g}{2} [t^{2} - {(t - Δ t)}^{2}]

= \frac{g}{2} (Δ t) (2 t - Δ t)

= \frac{g}{2} (\frac{1}{n}) (\sqrt{\frac{8 d}{g}} - \frac{1}{n})

Δ h = \frac{g}{2} (\sqrt{\frac{8 d}{n^{2} g}} - \frac{1}{n^{2}}) .

Commented by Tinkutara last updated on 07/Jun/17

Thanks Sir!