Question

A stone is dropped from the top of a tall cliff and n seconds later another stone is thrown vertically downwards with velocity u. Then the second stone overtakes the first, below the top of the cliff at a distance given by:

[Assume u sufficiently enough to overtake the first stone]

• A. $\left(\frac{g}{2}\right){\left[n\frac{(\frac{gn}{2}-u)}{gn-u}\right]}^2$
• B. $\left(\frac{g}{2}\right){\left[n\frac{(gn-\frac{u}{2})}{gn-u}\right]}^2$
• C. $\left(g\right){\left[\frac{(gn-u)}{gn-\left(u/2\right)}\right]}^2$
• D. $\left(\frac{g}{5}\right){\left[\frac{(gn-u)}{gn-\left(u/2\right)}\right]}^2$

Answer 1

The correct option is A $\left(\frac{g}{2}\right){\left[n\frac{(\frac{gn}{2}-u)}{gn-u}\right]}^2$

Let after $t$ seconds second stone overtake the first stone at distance $h$ from the cliff.

Using $s=ut+\frac{1}{2}g{t}^2$,

For the first stone,

$\Rightarrow -h=-\frac{1}{2}g(n+t{)}^2$ $\Rightarrow h=\frac{1}{2}g(n+t{)}^2$ ...1

For the second stone,

$\Rightarrow -h=-ut-\frac{1}{2}g{t}^2$ $\Rightarrow h=ut+\frac{1}{2}g{t}^2$ .....2

Equating equations 1 and , $\Rightarrow ut+\frac{1}{2}g{t}^2=d\frac{1}{2}g(n+t{)}^2$ $\Rightarrow t=\frac{g{n}^2}{2(u-gn)}$

Putting $t$ in equation 1, $\Rightarrow h=\left(\frac{g}{2}\right)[n+\frac{g{n}^2}{2(u-gn)}{]}^2$ $\Rightarrow h=\left(\frac{g}{2}\right)[n\frac{(\frac{gn}{2}-u)}{gn-u}{]}^2$