Question

A ball is dropped vertically from a height of h onto a hard surface. If the ball rebounds from the surface with a fraction r of the speed with which it strikes the latter on each impact, what is the n et distance travelled by the ball up to the $10\,th$ impact?

• A. $2h \left(\frac{1-r^{10}}{1-r}\right)$
• B. $h \left(\frac{1-r^{20}}{1-r^{2}}\right)$
• C. $2h \left(\frac{1-r^{20}}{1-r^{3}}\right)-h$
• D. $2h \left(\frac{1-r^{20}}{1-r^{2}}\right)-h$

Answer 1

Answer: (D)

Net distance travelled by ball is
$d=2h_{1}+2h_{2}+2h_{3}+\ldots$
or, $d=(\frac{\upsilon_{o}^{2}}{g}+\frac{\left(r\upsilon_{0}\right)^{2}}{g}+\frac{\left(r^{2}\upsilon_{0}\right)^{2}}{g}+ \ldots + $ upto $10$ terms $)-h$

$v_{0}=\sqrt{2 g h} $
$\Rightarrow \frac{v_{0}^{2}}{g}=2 h$
$\Rightarrow d=\frac{v_{0}^{2}}{g}\left(1+r^{2}+r^{4}+\ldots+10\right.$ terms $)-h$
$=2 h\left(\frac{1-\left(r^{2}\right)^{10}}{1-r^{2}}\right)-h$
$=2 h\left(\frac{1-r^{20}}{1-r^{2}}\right)-h$