A ball falls (initially at rest) from a height of 10m, and bounces repeatedly.

Question

On each successive bounce, the ball rises only 2/3 as high as on its previous bounce (due only to energy dissipation on impact). The vertical height attained, and the time required to complete each whole bounce (impact to impact) follow the relationship d = 1.25 t^2, where t is time is seconds, and d is height attained in m. What total time is required after release (rounded to the nearest 0.1 s if necessary) for the ball to impact the ground for the 4th time? (Assume the rebound from an impact is instantaneous.)

Answer 1

Strategy: Construct a table of heights and durations for each stage of the bounce set. Pay special attention to the initial drop, since it is “different” geometrically (only 1/2 of a bounce)!

Tactics: For a freely rising and falling object under the influence of gravity (neglecting air resistance), the paths upwards and downwards are reversible with respect to time. Therefore, for the initial drop, the time is only ½ that of a complete bounce rising to 10m height. Construct a table therefore:

Bounce # Height (d) attained t^2 t t (incremental)(sec)

0 (the initial fall) 10 8 (but, for “2t”) √2 (1/2 bounce) 1.4142

1 10*2/3 8*2/3 4√3/3 2.3094

2 10*(2/3)^2 8 * (2/3)^2 4√2/3 1.8856

3 10*(2/3)^3 8 * (2/3)^3 8√3/9 1.5396

total 7.1488 → 7.1 s

Note that intermediate roundings to less than 4 decimal places would have made it uncertain whether 7.1 or 7.2 s was the correct value as rounded! Also, since an exact rebound fraction can be carried through, it should be, as this makes intermediate calculations exact rather than subject to multiple round-off uncertainty.

As a useful mental check on the correct t calculations, you may note that t (bounces 1 and 3) are related by factor 2/3, as are 2*t(bounce 0) and t(ounce 2).