ERJ Brainteaser: January 2019

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.