Ball A is dropped from the top of a building. At the same instant another bowl B is thrown vertically upward from the ground.
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.
$\endgroup$3 Answers
$\begingroup$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 $$\frac{S_B}{S_A+S_B}={2 \over 3}$$
$\endgroup$ 5 $\begingroup$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
$$h=\frac23 H$$.
$\endgroup$ 2 $\begingroup$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 $0$.
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