Question

An object A is released from the top of a vertical building of height $H$. At an instant when it has descended $xm$ from the top, another object B is released from $ym$ below the top of that building. Right from this instant, both these objects take the same amount of time to reach the ground. Which of the following expressions in the given situation is correct ?

• A. $H=(x+y{)}^2/4x$
• B. $H=(x+y{)}^2/4y$
• C. $H=(x+y{)}^2/2x$
• D. $H=(x+y{)}^2/2y$

Answer 1

The correct option is A $H=(x+y{)}^2/4x$


Since acceleration is constant, we can use the equations for uniformly accelerated motion.

For A,
$u=0,S=x$

$v^2=u^2+2ax=2ax$
$v=\sqrt{2ax}$ is the velocity of A at point O.

Let the time taken for A to reach the ground from O be $t$

$s=ut+\frac{1}{2}a{t}^2$
$H-x=\sqrt{2ax}t+\frac{1}{2}a{t}^2$

For B,

$s=ut+\frac{1}{2}a{t}^2$
$H-y=0+\frac{1}{2}a{t}^2$
$\Rightarrow t=\sqrt{\frac{2(H-y)}{a}}$

Since time taken is the same, substituting in the first equation for A,

$H-x=\sqrt{2ax}\sqrt{\frac{2(H-y)}{a}}+\frac{1}{2}a\frac{2(H-y)}{a}$

$\Rightarrow H-x=\sqrt{4x(H-y)}+H-y$
$\Rightarrow (-x+y{)}^2=4x(H-y)$
$\Rightarrow y^2+x^2-2yx+4xy=4xH$

$\Rightarrow (x+y{)}^2=4xH$
$\Rightarrow H=\frac{(x+y{)}^2}{4x}$