Vhen a tennis ball is dropped, it bounces and then rises. he ball rises to \( 60 \% \) of the height from which it is dropped. he ball is dropped from a height of 2 metres. a) Calculate the height of the rise after the first bounce. (b) Calculate the height of the rise after the second bounce. The ball carries on bouncing, each time rising to \( 60 \% \) of the last rise. (c) For how many bounces does it rise to a height greater than 20 cm ? Show your working

**(a) Height after the first bounce** The ball is dropped from a height of 2 metres. After the bounce, it rises to \( 60\% \) of that height. Thus, the height after the first bounce is calculated as: \[ \text{Height} = 0.6 \times 2 = 1.2 \text{ metres.} \] --- **(b) Height after the second bounce** After the first bounce, the ball reaches \( 1.2 \) metres. It then rises to \( 60\% \) of the previous height on the second bounce: \[ \text{Height} = 0.6 \times 1.2 = 0.72 \text{ metres.} \] --- **(c) Number of bounces with a rise greater than 20 cm** Let the height after the \( n \)-th bounce be denoted by \[ h_n = 2 \times (0.6)^n. \] We are looking for the number of bounces such that \( h_n > 0.20 \) metres. Set up the inequality: \[ 2 \times (0.6)^n > 0.20. \] Divide both sides by 2: \[ (0.6)^n > \frac{0.20}{2} = 0.10. \] Taking the natural logarithm on both sides: \[ \ln \left((0.6)^n\right) > \ln(0.10). \] Using the logarithm property \( \ln(a^n) = n \ln(a) \): \[ n \ln(0.6) > \ln(0.10). \] Since \( \ln(0.6) \) is negative, dividing each side by \( \ln(0.6) \) reverses the inequality: \[ n < \frac{\ln(0.10)}{\ln(0.6)}. \] Calculating the values: \[ \ln(0.10) \approx -2.3026,\quad \ln(0.6) \approx -0.5108. \] \[ n < \frac{-2.3026}{-0.5108} \approx 4.507. \] Since \( n \) must be an integer, the ball will bounce to a height greater than 20 cm for \( n = 1, 2, 3, \) and \( 4 \). Thus, the ball rises to a height greater than 20 cm for **4 bounces**.