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 $10th$ 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
$d=2h_1+2h_2+2h_3+\dots$
or, $d=(\frac{{\upsilon }_o^2}{g}+\frac{{\left(r{\upsilon }_0\right)}^2}{g}+\frac{{\left(r^2{\upsilon }_0\right)}^2}{g}+\dots +$ upto $10$ terms $)-h$

$v_0=\sqrt{2gh}$
$\Rightarrow \frac{v_0^2}{g}=2h$
$\Rightarrow d=\frac{v_0^2}{g}(1+r^2+r^4+\dots +10$ terms $)-h$
$=2h\left(\frac{1-{\left(r^2\right)}^{10}}{1-r^2}\right)-h$
$=2h\left(\frac{1-r^{20}}{1-r^2}\right)-h$