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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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Ball A is dropped from top of a building. At same instant ball B is thrown vertically upwards from ground. When the balls collide, they were moving in opposite directions and speed of A is twice the speed of B. At what fraction of the height of building did the collision occur

\[(a)\;\frac{1}{3} \quad (b)\;\frac{2}{3} \quad (c)\;\frac{1}{4}\quad (d)\;\frac{2}{5}\]

Can you answer this question?
 
 

1 Answer

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Let h be total height let x be desired fraction.
For Ball A
$(1-x)h=\large\frac{1}{2} $$gt^2$
$=>t=\sqrt {\large\frac{2(1-x)h}{g}}$
For Ball B
$xh=ut-\large\frac{1}{2}$$ gt^2$
$=>xh=u\bigg(\sqrt {\large\frac{2(1-x)h}{g}}\bigg)$$-(1-x)h$
Solving
$u=\sqrt {\large\frac{gh}{2(1-x)}}$
$v_A=2v_B$ at the time of collision
${v_A}^2=4 {v_B}^2$
$2g(1-x)h=4 \{u^2-2gxh\}$
$(1-x)=\large\frac{1}{1-x}$$-4x$
$x=\large\frac{2}{3}$
Hence b is the correct answer.
answered Jul 1, 2013 by meena.p
edited May 24, 2014 by lmohan717
 

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