Physics Question

Let me apply my physics knowledge of kinematics to waterpolo.



Seng Kong wants to shoot a ball into the net during a waterpolo match. Assume that there is no air resistance and taking g as 10m/s. Assume all other relevant information.
Calculate the height he jumped.
Calculate the distance away from the goal post.
Calculate the time the ball took to travel to the goal post.
Calculate the angle he bounced the ball.
Calculate whether the ball is on target.
Calculate whether the ball will be a goal.

1. First we assume the depth of the swimming pool to be 1.8m and we set the reference point as the surface of the swimming pool.

Resolve vertically:
Swimming pool depth - sengkong's height=1.8-.175=0.05m
Sengkong's arm's length= 70cm
Sengkong's length strength = 100N
Distance = 100N/Sengkong's weight= 0.5 m

Thus sengkong jumped 0.5+0.05+0.7=1.25 metres.

2. Resolve horizontally,
Sx=Ux T

Since sengkong shoots the ball at a very high velocity which is difficult to calculate, the distance is unknown. However, a rough estimate will give you the answer to be (5.00+-0.1). This is because the pool is 25m and it is divided into 250 parts so the smallest division is 0.1m and that is the least uncertainty. 5m comes from the new waterpolo rules more commonly known as shooting paradise.

3. Resolve vertically,
Sy= UyT + 0.5 AyT

Sengkong's ball has very fast acceleration and very fast velocity. The graph sketch for sengkong's shots are all exponential graph. Sengkong's shots can be considered a many-to-one function as every ball that goes to him will leave his hands very fast. Taking K as a konstant, we can see that time is inversely proportional to his acceleration and velocity.

We can work out time to be 0.5sec because that is the minimum point of the graph f(x)=sengkong. If you dont believe me do plot it out in graphmatica.

4. The angle he bounced the ball can be calculated from the different distances. Using pythagoras theorem we can find x to be the root of 25+50. Using binomial theorem, we can find y to be around 10. Together using trigonometry we can calculate angle theta using the formula:

tan theta= X/Y

theta=50 degrees

Since angle of incidence = angle of relfection provided there is no defraction of the ball, sengkong's ball will bounce at an angle of 50 degrees.

5. Whether the ball is on target can be calculated by the following formula:

LUCK * FORM * Fatigue * Number of trainings attended the whole year/ number of trainings attended the whole year *100%

This is a very complicated formula and students are not required to know this. Thus the answer given will be 96.7% rounded to the least sf as this is a multiplication.

6. Whether the ball will be a goal can be calculated by the following formula:

Keeper's luck * form * leg strength/ total leg strength * calf muscles * thigh muscles=

Since the keeper was luckless that day, chances of sengkong scoring is zero.

I think the ball hit the keeper's face and bounce back. If you plot the displacement graph, you can notice that there is a inverse function. Because the ball went there and can also be mapped back.

I think my physics very powerful. PROVEN!