Last updated on 21 April/98
1. What is the launch velocity to jump reach a height of .8m?
v^2 = 2gh ;
v =sqrt(2gh) =sqrt(2*10* 0.8) = 4m/s
2. What is the 'hang time' for this jump?
for half the jump, h = 1/2 gt^2; t = sqrt(2h/g) = sqrt(2 *0.8/10)= 0.4 sec
hang time is twice this t = 0.8 seconds.
3. How long does the jumper stay above .4m?
To fall from the highest point by (.8 - .4) = .4 m, we have t/2 = sqrt(2 * 0.4/10)
= 0.28 sec. Hang time, t = 0.56 seconds. The majority of the hang time is spent at the upper half of the jump because the avg speed is the smallest.
4. If a defensive player has a delay of .2 second before jumping,
how much lower will she be of the offensive player [at her peak
of .8m] who jumped at t=0? [Assume both jump .8m]
The defensive player has 0.2 seconds to rise. This will put her at a height of
h = v0t - 1/2 gt^2 = 4*.2 - 1/2 * 10 *0.2^2 = .8 - .2 = 0.6m;
Since the second player is in the fast part of the jump, she is able to get up and make trouble for the offensive player in spite of jumping 0.2 seconds later. If the defensive player has a reach of 20cm more, she could block the shot.
5. Does Michael Jordan have a special dispensation to stay in the air longer than others for the same height of jump? He seems to do so? Name several reasons why it seems to be so. Argue why he MUST be in the air for the same time for the same height of jump.
He seems to hang in the air longer by the amount of smooth motions he makes while in the air to be able to get a shot cleanly off to the basket. Our perceptions of how long it would take mortals to make those motions so smoothly make us THINK that he's in the air for a long time.
Michael Jordan's body has the same gravitational acceleration as anyone else. For a given vertical height of jump, he would stay in the air as long as anyone or anything else. The only way that Jordan could stay longer in the air is to get some other force to work in his favour.
6. Dancers in the 'Jete' movement seem to hang in the air longer than others. There is an illusion of floating. Discuss why even Dancers can't float and why there is an illusion of floating. A famous dancer when asked how he did it remarked that " I just jump in the air... then I pause".
While the center of mass of a body follows a parabolic trajectory, the parts of the body may do something else. A dancer smoothly raises the extended legs during the jete which has the effect of raising the center of mass relative to the dancers head. When this is done well, the viewer watching head of the dancer will see that the head seems to float momentarily and interprets this as the dancer floating.
7. How long does a tennis ball stay within a target region of 4" if the tossed ball is just at the peak of it's arc?
At the peak, the tennis ball will fall 4" or 0.1 m in t = sqrt(2h/g) sqrt(2*.1/10)=0.14 sec
The total time [counting the rising and falling within this target region] = 0.28 seconds.which is plenty of time for a good strike.
8. How long does a tennis ball stay within a target region of 4" if the tossed ball is tossed high and falls one meter to reach the target region?
If you were to hit the ball on the way down after it has fallen for 1 meter, the velocity would be v=sqrt(2gh) =sqrt(2*10*1) = 4.5m/s
and the time in the target region would be t = y/v = 0.1/4.5 = 0.022 seconds which is more than 10x less than the top of the arc case. You would have to have superb timing to hit the ball well within this accuracy.
9. What is the center of mass of person? Is it always on a part of the body? Is it always on the same spot on your body? How can you raise the cm of your body relative to the cm position when you are standing erect.
The center of mass is that point at which you would balance in all directions. If you stand erect, the cm is above your navel and in the center of your body. If you were to bend and touch tour toes, the cm would shift to a point between your trunk and your hips which would be in the air AND NOT on your body. To raise your cm, while standing , you can raise your arms.
1. A long jumper approaches the takeoff board with a velocity of 10m/s. she is able to maintain this speed and still takeoff with a vertical velocity of 3.3m/s
a) What is the hang time for the cm to get back to the
original height?
the fall time can be calculated by understanding that the
fall and rise are just the T image of each other. v = gt for the
fall or t =v/g = 3.3/10 = 0.33 seconds. The total hang time is
0.66 seconds
b) How far does the jumper go during this time?
During this time the cm of the jumper will have gone horizontally by x = v0x*t = 10*0.66= 6.6m or 21.7 feet.
c) what is the strategy of the jumper to get a longer measured
jump?
To get the longest measured jump, the jumper should try to
get her mark on the sand as far forward as possible relative the
cm and to land with the cm as low as possible. This is usually
carried out by extending the legs in front of the sitting positon
of the body. The jumper must be careful not to make other marks
on the sand with other parts of the body.
d) How high does the cm of the jumper reach at the highest
point?
The maximum height of the jumper reaches is calculated from:
v^2=2gh. h = v^2/(2g) = 3.3^2/(2*10)= 0.5m or 20"
2. How fast must a ball be travelling in order for a pitched ball to be almost straight... as defined by the downward deflection is less than 1"?
To reach the plate 18m in a time for the ball to drop by 1" or 1/39m, the time must be:
t = sqrt(2h/g)= sqrt(2/39/10)=0.07 seconds
The horiz velocity must be v = 18/.07 = 257m/s = 591 mi/hr!!!