Now I have resolved this question a little in different ways. I did not arrive at the appropriate answer, so I have to understand what I have to fix.

You are watching: A ball is dropped from the top of a building

Let the building be of height H and let the balls collide at h and time t

Distance took a trip by A$$H-h=frac 12 gt^2$$So $$t^2=frac 2g (H-h)$$And distance spanned by B is

$$h=ut - frac 12 gt^2$$Subsisting the value of t$$h=usqrt frac2(H-h)g - (H-h)$$Squaring after doing appropriate simplification $$H^2=frac2u^2(H-h)g$$I don’t recognize how to continue additionally. Please assist.

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asked Sep 6 "19 at 16:44

Aditya Aditya

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## 3 Answers 3

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1

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The distances traveled by the two balls

$$h =ut - frac 12 gt^2$$$$H - h = frac 12 gt^2$$

Take their ratio

$$ frac hH-h = frac2ugt -1 ag1$$

Let $u_a$ and $u_b$ be the velocities at collision, then $gt$ is pertained to them via

$$gt = u-u_b = u_a ag2 $$

And, at collision,

$$u_a= 2u_b ag3$$

From (2) and also (3), we acquire $u=3u_b$, which leads to

$$gt = u-u_b = frac23 u$$

Plug over $gt$ into (1) to get

$$ frac hH-h = frac 21$$

Therefore, they collide at

$$h=frac23 H$$.

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edited Sep 7 "19 at 6:08

answered Sep 6 "19 at 17:32

QuantoQuanto

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2

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At first A is dropped at remainder , after time t it"s velocity is $2v=gt$ , or $v=5t$. and also let B is thrown by Initial velocity u . $u-gt=v$ or $u=15t$ . currently use $s= ut+1 over 2gt^2$.

$$S_A=0(t)+5t^2=5t^2$$$$S_B=15t(t)-5t^2=10t^2$$$$S_A over S_B=1 over 2 $$&$$S_A+S_B=H$$>Ratio is $$fracS_BS_A+S_B=2 over 3$$

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edited Sep 6 "19 at 17:22

answered Sep 6 "19 at 16:58

RishiRishi

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The closing rate will certainly reprimary consistent and is equal to the initial rate of $B (Vib)$ so $Va + Vb = Vib$.

When they collide, $Va = 2Vb$

We can check out that $Vib = 3Vb$ at the collision.

So: The rate of A goes from $0$ to $2Vb$ and also the rate of B goes from $3Vb$ to $Vb$

Distance is average velocity times time.

$s(Va) = Vbcdot t$

$s(Vb) = 2Vbcdot t$

$B$ travels twice as much as $A$ so the fraction of the height the collision occurs is $frac23$.

I think tright here is one more solution wbelow $B$ reverses direction and starts to fall and is eventually recorded by $A$.

Here $Va - Vb = Vib$

And $Vib = Vb$ at the collision.

In this scenario, the speed of $A$ goes from $0$ to $2Vb$ and also the speed of $B$ goes from $Vb$ to $-Vb$.

$s(Va) = Vbcdot t$

$s(Vb) = 0cdot t$

In this scenario the height the collision occurs is $0$.

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edited Sep 6 "19 at 18:12

answered Sep 6 "19 at 17:18

Phil HPhil H

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