Question

A steel ball is dropped on a hard surface from a height of 1 m and rebounds to a height of 64 cm. The maximum height attained by the ball after $n^{th}$ bounce is (in m)-

• A. $(0.64{)}^{2n}$
• B. $(0.8{)}^{2n}$
• C. $(0.5{)}^{2n}$
• D. $(0.8{)}^n$

Answer 1

Answer: (B)

$(0.8{)}^{2n}$

Let $v_n=$ velocity after $n^{th}$ bounce
$h_n=$ height after $n^{th}$ bounce
$h_0=1m=100$ cm
$h_1=64$ cm
$e=$ velocity of separation/velocity of approach$=\frac{h_1}{h_0}$
$v_0^2=2g{h}_0$
$v_1^2=2g{h}_1$

$\frac{h_1}{h_0}=\frac{v_1^2}{v_0^2}$;

$\frac{h_2}{h_1}=\frac{v_2^2}{v_1^2}$;
$\frac{h_n}{h_{n-1}}=\frac{v_n^2}{v_{n-1}^2}$;
$\Rightarrow$; $\frac{h_n}{h_0}=e^{2n}$;

$e^2=\frac{64cm}{1m}=0.64={0.8}^2$

$\Rightarrow$; $\frac{h_n}{h_0}=(0.8{)}^{2n}$