When the balls collide, they are moving in oppowebsite direction and the rate of A is twice the speed of B. At what fractivity of the height did the collision occur?

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 \$endgroup\$

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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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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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