A tennis ball is hit vertically upwards at a speed of 12 m s−1 from a point 60m above the ground. Find: a) the speed of the ball when it strikes the ground; and b) the total time the ball is more than 64m above the ground.

**Answer**: Quite a few single and double faults for this testing teaser, so extra well done to the trio who aced it (see solutions below): **Michele Girardi**, quality manager, Scame Mastaf Spa, Suisio, Italy; **John Bowen**, rubber industry consultant, Bromsgrove, Worcs, UK; **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands.

Solutions

*Andrew Knox*

*All you need to know here is that under the acceleration of gravity g, velocity v = g.t, and distance travelled d = 1/2.g.t.t (that is t squared). *

*Assuming negligible air resistance effects, the tennis ball decelerates on its way up in the same way it will accelerate on its way down.*

*a) As velocity v = g.t, time t on way up to decelerate from speed 12 m/s to 0, or on way down from 0 to 12 m/s, is v/g = 12/9.81 = ca. 1.223 s.*

*In this time the ball has travelled distance d = 1/2.g.t.t = 9.81/2 x 1.223 x 1.223 = ca. 7.34 m.*

*So, maximum height reached will be 60 + ca. 7.34 = ca. 67.34 m.*

*When the ball falls from this height, time t to hitting the ground is therefore given by: t squared = 2 d/g = 2 x 67.34/9.81 = ca. 13.7287, or t = ca. 3.705 s.*

*Velocity at impact v = g.t = 9.81 x 3.705 = ca. 36.35 m/s.*

*b) Height travelled by the ball above 64 m is ca.3.34 m. Time spent above this height is therefore twice the time it takes to fall from the maximum height (v=0) to 64 m.*

*Time to fall 3.34 m is given by t = square root of 2.d/g = root 6.68/9.82 = root 0.68, so t = ca. 0.825 s.*

*So, total time spent above 64 m = 2 x 0.825 = ca. 1.65 s.*

*John Bowen*

*This one requires use of the Equations of Motion and a bit of algebra.*

*To calculate the first answer, we need to calculate the ultimate height from which the ball fell - this is where it originally stopped before accelerating earthwards and assumes no effect from air resistance. *

*We use vsqd - usqd = 2fs where v = 0 [stopped] u = 12 m/s f = acceleration due to gravity [-9.81m/s/s] and s = distance*

*so 0 - 144 = 2 x- 9.8 x s, so s = 7.35m. As the ball started 60m above the ground, total height it falls through = 67.35m.*

*Again, using the above formula, from a standstill, vsqd = 2fs, vsqd = 2 x 9,8 x 67.35 = 1320.06, so v = 36.33 m/s*

*To calculate the time above 64m above the ground, or 4m above its starting point, we need the formula s = ut + 1/2.f.tsqd*

*so 4 = 12t + 1/2.-9.8.tsqd which rearranges to: 4.9.tsqd - 12t +4 = 0*

*Solving for t gives values of 0.4 or 2.05 seconds, which means the ball passes the 64 m height upwards at 0.4 seconds, then downwards at 2.05, *

*So it is above 64m high for 1.65 seconds.*

*Michele Girardi*

*The equation of motion applicable are*

Y=Y0+V0*t-1/2.g.t^2 V=V0-g.t

*substituting Y0=60 g=9.8 V0=12*

* Y=60+12*t-4.9*t^2*

*To avoid some boring algebra, we can use an excel file and the solver function to find the solutions for Y equal to 64,64 and zero*

*point t Y V*

start 0,0 60,0 12,0

cross 64+ 0,4 64,0 8,1

cross 64- 2,1 64,0 -8,1

ground 4,9 0,0 -36,3

*delta time 1,7*