At a tennis ball manufacturing company, materials scientists are testing their latest product range for bounce.
In one experiment, a ball is thrown vertically upwards with a speed of 16 m s−1 from a point 80 metres above the ground.
The test guys have to work out the speed with which the ball strikes the ground as well as the total time the ball is more than 85 metres above the ground.
When it lands, the ball rebounds with 20% of the speed with which it strikes the ground. The team then try to work out the greatest height reached by the ball when it rebounds.
What should be the results of the three tests?l
Answer: This really was one of our tougher teasers, so hats off to everyone who had ago. And, extremely well done to the following ace readers, who skilfully worked out the answers at: around 43m s−1, 2.6s, and 3.7m (solutions below): Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; and Michele Girardi, Scame Mastaf Spa, Suisio, Italy.
Solutions
John Bowen
Mr Newton comes in handy here, as we need to use the equations of motion where acceleration due to gravity is taken as 9.81ms-2
To calculate the speed the ball hits the ground, we need to calculate the initial height above the 80m start:
v.sqd-u.sqd =2fs: at the top velocity = 0, so 16 sqd = 2fs; 256 = 2*9.81*s = 13.05m.
We use this equation again downwards to the ground to calculate velocity at impact: v. sqd = 2fs
v.sqd = 2*9.81*93.05 [total fall is 80 + 13..05 above] = 1825.64
v = 42.73ms-1 This is the velocity with which the ball hits the ground
Use the same equation to calculate the rebound height: vsqd - usqd = 2fs; u = 8.55 [20% of impact speed], v = 0, f = -9.81
73.1 = 2*9.81*s ; s = 3.73m
To calculate the time above 85metres we use the Equation of motion s = ut + 1/2 f.tsqd and solve the resulting quadratic equation using the 80m starting point and height [s] of 5 m:
5 = 16t - 1/2*9.81* tsqd
Rearranging, 4.9*tsqd -16t + 5 = 0
Applying the quadratic formula [x = [-b+/-{sq rt [bsqd - 4ac]}/2a]
t = 16 +/-SqRt[256-98]/9.8 = [16+/- 12.6]/9.8
t = 0.35 [upwards] and 2.92 [downwards]
So time above 85m high = 2.57 seconds
Michele Girardi
Y=Y0+V0*t-1/2.g.t^
Y0=80
g=10
V0=16
Y=80+16*t-5*t^2
dY/dt = 16-10*t
Using an excel file and the solver function to find Y equal to 85,85,0
point t Y dY/dt
start 0,00 80,00 16,00
cross 85 0,35 85,00 12,49
max 1,60 92,80 0,00
cross 85 2,85 85,00 -12,49
ground 5,91 0,00 -43,08
rebound V0' = 43.08*0,2 = 8,62
after rebound
V^2=V0'^2-2.g*Y
V^2=8.62^2-20*Y
Y =(0-8.62^2)/(-20)
max Y= 3.72m
Speed when the ball hits the ground : 43.1 m/s
Time above 85 m : 2.85-0,35 = 2.5s
Heigth of rebound 3.72 m
Bonus question
What should the values of X be in the third row of this table?
The key to the equally tricky Bonus question was to identify that adding the numbers in the first two squares if each row (right to left) and then adding the digits to get the next number, ie in the third column. So, in the fourth row, 8+8=16, 1+6 =7 and 8+7=15, 1+5=6. Very well done here to: Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; and Mehmet Koral, Erhardt-Leimer representative for Turkey, managing director, C&C Endüstriyel Danışmanlık, Eğitim ve Mümessillik Ltd, Göztepe, Istanbul, Turkey.
New teaser on Monday
