Section 1 (Answers could very) 1. The purpose of the study was to ...

Section 1
(Answers could very)
1. The purpose of the study was to examine reaction forces transmitted to the upper extremities
of high-level gymnasts during the round-off phase of the Yurchenko vault.
2. The two conditions were whether the elite-gymnast performed a Yurchenko vault, or a floor
exercise pass.
3. The two dependent variables were the peak vertical and anterior-posterior reaction forces.
4. High reaction forces transmitted to the upper-extremities are more dangerous than high
reaction forces transmitted to the lower-extremities because the lower-extremities are designed
for such weight bearing activities: the upper-extremities only involve relatively small bones of
the hand, wrist, and arm, with little cushioning; also the upper-extremities have less muscle
mass, relative to the lower-extremities.
Section 2
A. A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60
m/s. A 15 m/s wind is blowing in a northwesterly direction (45 degrees west of North).
1. Considering the initial velocity of the ball and the velocity of wind, what will be the
resultant velocity (m/s) of the golf ball assuming no other forces are acting on the ball?
First, draw a picture and combine the two velocity vectors using the tip to tail method.
15 m/s
Resultant
velocity
60 m/s
Next, resolve the 15 m/s velocity vector into vertical and horizontal components using sin and cosine.
Sin(45)*15 = 10.6
10.6
15 m/s
Cos(45)*15 = 10.6
10.6
Now, add the two vertical components together (10.6 and 60) and combine the new vertical vector
(70.6) with the resolved horizontal vector (10.6) using the tip to tail method.
10.6
Resultant
velocity
70.6 m/s

Next use Pythagoreans equation to solve for the resultant velocity
c = 71.4 m/s
2. What will be the resultant direction (º west of North) of the golf ball?
Use inverse tangent to solve for the resultant direction (angle).
tan -1 (10.6/70.) = 8.54 ° West of North
B. An 80-kg firefighter slides down a fire pole. After 1.3 seconds of sliding, the firefighter is
sliding at a velocity of 6.5 m/s, straight down the pole. The firefighter then grips the pole tighter
so that the force of friction exerted by the firefighter's hands on the pole is equal to the force of
gravity.
1. At this point (remembering Newton's Second Law), when the friction and gravitation
forces are equal, what is the downward acceleration of the firefighter?
No calculations are need here. Newton’s Second Law states that ∑F = m·a. So if the two vertical
forces are equal and opposite, the sum of the forces is equal to zero. Therefore, the acceleration
is also zero.
2. Knowing that acceleration indicates a change in velocity, if the firefighter continues to
hold his same grip, maintaining the same friction force, for another second what will be
his velocity?
If the acceleration is equal to zero, that would indicate there is no change in velocity. If there is
no change in velocity over time, the velocity of the firefighter would be the same as it was
previous. Therefore, the velocity remains to be 6.5 m/s.