Browse Questions

# The distance through which a body falls in the nth seconds is h. The distance through which it falls in the next seconds is

$\begin{array}{1 1} h \\ h+\large\frac{g}{2} \\ h-g \\ h+g \end{array}$

Answer : $h+g$
$y= \large\frac{1}{2}$$g(n+1)^2-\large\frac{1}{2}$$gn^2$
$\quad= \large\frac{R}{2} $$[ (n+1)^2 -n^2] \quad= \large\frac{R}{2}$$(2n+1)$
Also, $h= \large\frac{g}{2} $$(2n-1) \therefore y= \large\frac{g}{2} \bigg[ \large\frac{2h}{g}$$ +1+1 \bigg]$
or $y= h+g$