Ball A is dropped from the top of a building. At the same instant another bowl B is thrown vertically upward from the ground.

Question

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

Now I have solved this question a bit differently. I did not arrive at the right answer, so I need to know what I have to fix.

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

Distance travelled by A $$H-h=\frac 12 gt^2$$ So $$t^2=\frac 2g (H-h)$$ And distance covered by B is

$$h=ut - \frac 12 gt^2$$ Subsisting the value of t $$h=u\sqrt {\frac{2(H-h)}{g}} - (H-h)$$ Squaring after doing appropriate simplification $$H^2=\frac{2u^2(H-h)}{g}$$ I don’t know how to proceed further. Please help.

Rishi · Answer 1 · 2019-09-06T17:22:52.510

2

Initially A is dropped at rest , after time t it's velocity is $2v=gt$ , or $v=5t$. and let B is thrown by Initial velocity u . $u-gt=v$ or $u=15t$ . now use $s= ut+{1 \over 2}gt^2$.

$$S_A=0(t)+5t^2=5t^2$$ $$S_B=15t(t)-5t^2=10t^2$$ $${S_A \over S_B}={1 \over 2} $$&$$S_A+S_B=H$$ ] Ratio is $$\frac{S_B}{S_A+S_B}={2 \over 3}$$

edited Sep 06 '19 at 17:22

answered Sep 06 '19 at 16:58

Rishi

998

No I can’t. I don’t know in what context you are talking in – Aditya Sep 06 '19 at 17:09
since $v=at$ here a=g=10 (fix). Initially and after time t consider velocities. – Rishi Sep 06 '19 at 17:14
Okay so $S_a/S_b = \frac 12$ What did u do next? – Aditya Sep 07 '19 at 10:57
You can assume $S_A=x$ , then $S_B=2x$ and $S_A+S_B=3x$ , that's why – Rishi Sep 07 '19 at 12:22
That’s an interesting solution. Never thought it could be done that way. You should have put this in the answer though, it’s a little hard to follow without any (you guessed it!) context. – Aditya Sep 07 '19 at 12:32

Quanto · Accepted Answer · 2019-09-07T06:08:42.893

1

The distances traveled by the two balls

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

Take their ratio

$$ \frac {h}{H-h} = \frac{2u}{gt} -1\tag{1}$$

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

$$gt = u-u_b = u_a\tag{2} $$

And, at collision,

$$u_a= 2u_b\tag{3}$$

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

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

Plug above $gt$ into (1) to get

$$ \frac {h}{H-h} = \frac 21$$

Thus, they collide at

$$h=\frac23 H$$.

edited Sep 07 '19 at 06:08

answered Sep 06 '19 at 17:32

Quanto

97,352

I didn’t why $gt=u-u_b=u_a$ – Aditya Sep 07 '19 at 09:29
It is the equation for velocity change due to gravity for both balls. Ball A accelerates from 0 to $u_a$ and ball B decelerates from u to $u_b$ – Quanto Sep 07 '19 at 10:46

Phil H · Answer 3 · 2019-09-06T18:12:51.550

The closing speed will remain constant and is equal to the initial speed of $B\ (Vib)$ so $Va + Vb = Vib$.

When they collide, $Va = 2Vb$

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

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

Distance is average velocity times time.

$s(Va) = Vb\cdot t$

$s(Vb) = 2Vb\cdot t$

$B$ travels twice as far as $A$ so the fraction of the height the collision occurs is $\frac{2}{3}$.

I think there is another solution where $B$ reverses direction and starts to fall and is eventually caught 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 the speed of $B$ goes from $Vb$ to $-Vb$.

$s(Va) = Vb\cdot t$

$s(Vb) = 0\cdot t$

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

Ball A is dropped from the top of a building. At the same instant another bowl B is thrown vertically upward from the ground.

3 Answers3