Ball A is dropped from the top of a building. At the same instant another bowl B is thrown vertically upward from the ground.
Solution 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
$$h=\frac23 H$$.
Solution 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 $$\frac{S_B}{S_A+S_B}={2 \over 3}$$
Related videos on Youtube
Aditya
Updated on August 01, 2022Comments
-
Aditya over 1 year
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.
-
Aditya about 4 yearsNo I can’t. I don’t know in what context you are talking in
-
Rishi about 4 yearssince $v=at$ here a=g=10 (fix). Initially and after time t consider velocities.
-
Aditya about 4 yearsI didn’t why $gt=u-u_b=u_a$
-
Quanto about 4 yearsIt 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 $u_b$
-
Aditya about 4 yearsOkay so $S_a/S_b = \frac 12$ What did u do next?
-
Rishi about 4 yearsYou can assume $S_A=x$ , then $S_B=2x$ and $S_A+S_B=3x$ , that's why
-
Aditya about 4 yearsThat’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.