In tests for the next World Cup, a football is modelled as a particle and air resistance is ignored. A player kicks the ball from a height of 0.6m above the ground, propelling it vertically upwards at a speed of 10.5m/s. From the modelling, what is the greatest height above the ground reached by the ball, and calculate the length of time the ball is more than 2m above the ground.

**Answer**: Said it might get tougher and it did with only three correct replies to this week’s teaser, from: **John Bowen**, consultant, Bromsgrove, UK: **Bharat B Sharma**, Sr VP product development & technical service (elastomers), Reliance Industries Ltd, Gujarat, India; **Paul Knutson**, textile engineer, Timken Belts, Springfield, Missouri, USA. Their workings out of the correct answers **6.22 metres** and **1.86 seconds** are each given below:

**John Bowen**

*This problem uses the basic Laws of Motion:*

*i] To calculate height reached – ie when the ball stops in its vertical trajectory we use*

*v sqd – u sqd = 2fs, where f = acceleration due to gravity [ in this case -9.81m/sec/sec] and s = distance, u = initial velocity, v = final velocity, so*

*0 – 110.25 = -2*9.81*s*

*s = 5.62, so total height = 5.62 + 0.6 = 6.22m*

*ii] to determine the time above 2m we need to solve a quadratic of the form s = ut + 1/2f*t sqd where t = time in seconds; s = 1.4 [as it starts at a height of 0.6m]*

*so 1.4 = 10.5*t – 1/2*9.81*t sqd*

*solving for t gives values of 1.99 and 0.143, so the ball is above 2m height for 1.85 seconds*

*Greatest height above the ground reached by the ball = 6.225m*

*Total time the ball is more than 2m above the ground.= 1.857 Sec. (**gravity considered as 9.8m/sq.second)*

**Bharat B Sharma **

*1 Initial height = 0.6 *

*v2 = u2 + 2as v=0 *

*u=10.5m/s *

*(10.5)2= 2×9.8xs a= (-9.8m/sec2) *

*s= (10.5×10.5/2×9.8 )+0.5 m(Initial height) *

*s=5.625m+0.6m **= 6.225m *

*2. Time require to keep ball above 2 m from ground (upward 2m to 6.225m and return)*

*Distance covered 4.225 m upward and 4.225 m down ward *

*Going up time Final speed = 0, s= 4.225 m, a =-9.8m/sq.sec== *

*Initial speed at 2 m = sq root of 2×9.8×4.225= sq rt of 82.81 m/sec*

* =9.1 m/sec *

*Time taken to trave 4.225 m= (V=u+at) =0=9.1-9.8t) *

* = time = 9.1/9.8= 0.9286 sec *

*Going down = distance covered 4.225m, Initial speed is 0 and a=9.8m/sec2*

*v sqare=u sq+2as u=0, a=9.8m/sq.sec, s=4.225) *

*v sq= 82.81 *

*v 9.1m/sec *

*Time for return journey will also be 0.9286sec *

*Total time the ball will be above 2 meter height =2*0.92857= 1.857 sec.*

**Paul Knutson**

*Greatest height = 6.22 m. Time above 2 m is 1.86s. All derived from formula y=v0*t-.5*g*t^2*

Find the next number in the following sequence: 10, 9, 17, 50, 199, _

**Answer**: Clues were offered but not really needed as correct replies rolled in, including Michele Girardi’s a(i+1)=a(i)*i-1 making the sequence: 1,10; 2,9; 3,17; 4,50; 5,199; 6, **994**. Well done in order of reply to: **John Droogan**, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; **Andrew Beasley**, product analyst, Hankook Tyre UK Ltd, UK; **John Bowen**, consultant, Bromsgrove, UK: **Dave Stuckey**, Dow, South Wales, UK; **Hans-Bernd Lüchtefeld**, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; **Bharat B Sharma**, Sr VP product development & technical service (elastomers), Reliance Industries Ltd, Gujarat, India; **Michele Girardi**, Scame Mastaf Spa, Suisio, Italy; **Michael Easton**, sales and marketing director, Globus Group, Manchester, UK; **Jon Cutler**, materials development manager, Trelleborg Sealing Solutions, Tewkesbury, UK; **Martin Jones**, customer service manager, network logistics & transport, DHL Supply Chain UK; **Amparo Botella**, Ismael Quesada SA, Spain; **Thierry Montcalm, **R&D manager, Soucy Techno Inc., Canada; **Trey Thies**, growth strategist, engineered performance products, Milliken & Co., Spartanburg, South Carolina, USA; **Yuichi (Joe) Sano**, Sumitomo Electric Industries Ltd, Itami, Japan; **David Mann**, manager rubber technology, SI Group, Béthune, France.

Find the rubber connection in: England, Peru, Denmark, Malaysia.

**Answer**: A relatively easy one to start the year, as long you were not thrown off track by ‘Malaysia’ – things might get tougher as we go along. Well done to the following readers who identified EPDM in the country names: **David Mann**, manager rubber technology, SI Group, Béthune, France; **Jose Padron**, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; **John Droogan**, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; **Paul Knutson**, textile engineer, Timken Belts, Springfield, Missouri, USA; **John Bowen**, consultant, Bromsgrove, UK: **France Veillette**, chef environnement, usine de Joliette, Bridgestone Canada Inc. Canada; **Thierry Montcalm**, R&D and innovation manager, Soucy Techno, Canada; **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands.