1. FINAL REPORT FOR FORCE-PLATE LAB TEST GROUP 4 -15 SEPTEMBER
2011
Students:
Stavros Litsos
Emiliano Verzilli
ANSWERS TO THE QUESTIONS
1. Calculate the average weight of the quiet standing trial.
1. The average value of the quiet standing trial is the sum of all measurements
divided by the number of intervals and is 575,22 (N). This is also the average
value of the weight.
2. Create one figure showing 3 graphs: vertical GRF during quiet
stance, during walking and during running. Discuss interesting
features about these graphs.
2. See "3 graph=stance, running, walking" sheet.
By looking at the 3 graphs we notice immediately that the slope representing
the stepping moment is more rounded in the quiet standing trial than in the
next two graphs; the reason for this is that the time the force is applied to the
plate is longer in the quiet standing trial than in the following two tests.
Another noticeable aspect of these graphs is that the slope becomes
progressively steeper from the standing trial graph to the running graph; the
reason for this is that the magnitude of the force applied to the plate
increases.
Finally, we can point out that the running graph has one peak while the
walking one has two. This is due to the fact that the walking graph shows both
the heel and the toe touch. On the other hand, the running performance is
represented by one peak due to the velocity of the performance. What it
seems like a second peak in our graph is only due to the instability related to
the shoes used by our performer.
3) Create two graphs of Ground Reaction Force (GRF) vs Time (one for
each jump Draw and label lines on the graphs to represent the
following:
1. Body Weight line
2. Takeoff (TO)
3. Landing (L)
4. Flight Time (FT)

2. 3. See "countermove" sheet and "squat" sheet for answers to questions 3.1-
3.4)
3.5) Impulse up to the point of takeoff?
Countermove jump
ANSWER -168,37 (Ns) WITH TAKE OFF TIME AT 3,563 s found graphically
(THE SUM OF ALL IMPULSES FOR INTERVALS UP TO TAKE OFF)
Squat
ANSWER -424,31 (Ns) WITH TAKE OFF TIME AT 3,66 s found graphically
(THE SUM OF ALL IMPULSES FOR INTERVALS UP TO TAKE OFF)
3.6) Take off velocity ?
Countermove jump
ANSWER 2.87 m/s
(WE USE THE IMPULSE LAW F*t=m*v; first we calculate the mass m from the
average weight w= 575,22 (N) and the acceleration of gravity and we obtain
m= 58,63 kg; then we know that F*t at take off is -168,37 Ns, therefore v=
F*t/m = 2.87 m/s)
Squat jump
ANSWER 7.23 m/s
(WE USE THE IMPULSE LAW F*t=m*v; first we calculate the mass m from the
average weight w= 575,22 (N) and the acceleration of gravity and we obtain
m= 58,63 kg; then we know that F*t at take off is -424,31 Ns, therefore v=
F*t/m = 7,23 m/s)
3.7) Flight time?
Countermove jump
ANSWER 0,58 s
(WE USE ONE OF EQUATION OF CONSTANT ACCELERATION ;
the jumper is only affected by gravity when airborne; his take off velocity is
known and set at 2,87 m/s; his velocity at the apex of his flight is zero;
therefore we can use the equation 0=v+gt, 0=2,87m/s - 9,81m/s2 * t, t= 0,29
s; since the projection height and landing height are equal the time it takes to
reach the apex is one half of the total flight time and FT=2t=0,58 s).
Squat jump
ANSWER 1.52s
(WE USE ONE OF EQUATION OF CONSTANT ACCELERATION ;

3. the jumper is only affected by gravity when airborne; his take off velocity is
known and set at 7,5 m/s; his velocity at the apex of his flight is zero;
therefore we can use the equation 0=v+gt, 0=7,2m/s - 9,81m/s2 * t, t= 0,73
s; since the projection height and landing height are equal the time it takes to
reach the apex is one half of the total flight time and FT=2t=1.46s).
3.8) Flight height?
Countermove jump
ANSWER 0,41 m
(same reasoning as before but this time we need to find the maximum height;
therefore we use the equation 0=v12+2gh with v1=2,87m/s and g=-9,81m/s2;
we find that h= 0.41 m)
Squat jump
ANSWER 2,66 m
(same reasoning as before but this time we need to find the maximum height;
therefore we use the equation 0=v12+2gh with v1=7,23m/s and g=-9,81m/s2;
we find that h= 2.66 m)
Answer in a few words the following questions:
1. What does a longer flight time imply about jump height?
1) the jump height is higher
2. Compare the calculated and measured (from the graph) flight time.
What might be the reason for differences?
Countermove: FT from graph is 0,62 s; calculated FT is 0,58 s.
Squat: FT from graph is 0,59 s; calculated FT is 1,46s.
The reason for differences might be approximation related to measuring
directly from the graph.
3. Compare the force-time curves for the two vertical jumps. What
might be the reason for the differences in jump height?
The following explanation is based on both mechanical and physiological
reasons.
In the countermoving jump the total mechanical work performed is greater
than in the squat jump.
How high a person can reach after a jump, depends on the Kinetic Energy at
take-off which is equal to the mechanical work carried out during the jump
(see principle of work and energy).

4. The mechanical work (F.s) in a countermoving jump is greater because F (the
muscular force) is greater and this is due to two important physiological
reasons:
- the ability to store elastic energy in the eccentric phase increases;
- stretch reflexes leading to the activation of higher-lying motor units are
activated.
See Hills curve.
As a result the Kinetic Energy Ek=1/2mv2 at take-off increases; because the
mass stays the same it is v2 that increases, in other words the take-off
velocity increases.
Because the jump height is given by the equation h=v2/2g, we see that as the
Kinetic Energy increases so does the jump height.
(NB: the above equation for the jump height comes from the principle of
conservation of mechanical energy applied to the flight; we have that the
Energy at take off is equal to the Energy at the peak;
that is Ek + Ep at take off = Ek+Ep at peak; that is 1/2 mv2 + mgh at take off
= 1/2mv2 + mgh at peak;
then we take away m from both sides of the equation and set v at peak to 0 so
that we only have
1/2 v2 + gh at take off = gh at peak,
gh at peak - gh at take off= 1/2 v2 ,
h at peak - h at take off=v2/2g,
h=v2/2g)
NB: we are actually getting higher values for FT, flight height and Take
Off Velocity for squat jump rather than countermove. The reason to
this unusual results ia at present uknown to us.
Besides, we want to point out that there was an error in the execution
of the countermove jump because the performer swung his arms back
rather than keeping them on the hips.