Hi J, Wouldn’t it be a much faster way to calculate by simply looking at the problem as:
=> 2*(GP) – (Drop height)

Going upwards and downwards are the exact same GP, barring the first entry of the ball going upwards to drop height. Helps not calculating a new GP and rather simply multiplying the first GP with 2?

Yes and there is a still faster way too which avoids the GP totally. Which is why I am doing two posts on this particular question ðŸ™‚ Will cover the “even faster” methods in Wednesday’s sequel “Bouncing Ball – Endgame”.

Hi J, Wouldn’t it be a much faster way to calculate by simply looking at the problem as:

=> 2*(GP) – (Drop height)

Going upwards and downwards are the exact same GP, barring the first entry of the ball going upwards to drop height. Helps not calculating a new GP and rather simply multiplying the first GP with 2?

Yes and there is a still faster way too which avoids the GP totally. Which is why I am doing two posts on this particular question ðŸ™‚ Will cover the “even faster” methods in Wednesday’s sequel “Bouncing Ball – Endgame”.

regards

J