Question

A ball is thrown vertically upwards from the ground. It rises to a height of 10m and then falls and bounces. After each bounce, it vertically 2/3 the height of which is fell.
i) Find the height to which the ball bounces after the nth impact
ii) Find the total distance travelled by the ball from the first throw to the nth impact with the ground​

Answer 1

${\huge{\fcolorbox{yellow}{red}{\orange{\boxed{\boxed{\boxed{\boxed{\underbrace{\overbrace{\mathfrak{\pink{\fcolorbox{green}{blue}{Answer}}}}}}}}}}}}}$

(i)

$\sf{a_n = 20 \times {( \frac{2}{3} )}^{n - 1} }$

(ii)

$\sf S_n = 60 \{1 - { \frac{2}{3}}^{n} \}$

Step-by-step explanation:

$\underline\red{\textsf{Given :-}}$

height of ball (a) = 10m

fraction of height decreases by each bounce (r) = 2/3

$\underline\pink{\textsf{Solution :-}}$

(i) We will use here geometric progression formula to find height an times

${\blue{\sf{a_n = a {r}^{n - 1} }}} \\ \sf{a_n = 20 \times { \frac{2}{3} }^{n - 1} }$

(ii) here we will use the sum formula of geometric progression for finding the total nth impact

$\orange {\sf{S_n = a \times \frac{(1 - {r}^{n} )}{1 - r} }} \\ \sf S_n = 20 \times \frac{1 - ( { \frac{2}{3} })^{n} }{1 - \frac{2}{3} } \\ \sf S_n = 20 \times \frac{1 - {( \frac{2}{3}) }^{n} }{ \frac{1}{3} } \\ \sf S_n = 3 \times 20 \times \{1 - ( { \frac{2}{3}) }^{n} \} \\ \purple{\sf S_n = 60 \{1 - { \frac{2}{3} }^{n} \}}$