Hi everyone, my name is Tze Wei Lim and i am a exercise science undergraduate from Deakin University. I am currently undergoing my Master in Exercise Science (Strength and Conditioning) at ECU, and am having some problems with questions from my biomechanics assignment. I was wondering if anyone here could help me out with some of the questions that i have difficulties with ?

1. Neglecting the height of release, a baseball is thrown vertically upwards at 20 m/s, find:

a) vertical velocity after 2 s

b) maximum height

c) total flight time

d) time (seconds) that the projectile is at height 10.5 m.

What would happen to the values obtained in question 1 if the baseball was thrown at an angle of 30 degrees to the ground? (circle one option)

a) Increase, decrease, no change

b) Increase, decrease, no change

c) Increase, decrease, no change

2. At the instant an athlete is cutting, he is applying 200 N of medio-lateral force and 400 N of anterior-posterior propulsive force. Neglect vertical force in this problem.

a. Based on your knowledge of 3-D axis. Draw the how these two forces would be represented on a diagram, using both direction and magnitude.

b. Calculate the resultant horizontal force of this athlete. Resultant horizontal force = √((400N)2 + (200N)2) = √(160000+40000) = 447.21N (Solved!)

c. Calculate and describe the angle the force is being applied (you may show this on the diagram)

4. Show your work for the following and calculate what happens if an athlete is pulling a sled angle attached to his hip, resulting in an angle of force of pull at 15 degrees to the ground. If the coefficient of sliding friction between the sled and the grass is 0.8 and the sled has a mass of 25 kg, what is the acceleration of the sled if he is constantly pulling with a resultant force of 300 N? (this is a multi-step problem).

So far, i have managed to calculate these steps:

25kg * 9.81 = 245.25N

300sin15° = 77.65N (Vertical component)

245.25N – 77.65N = 167.6N (Reaction force)

300cos15° = 289.78N (Horizontal component)

F = μR, F = 0.8 * 167.6N, F = 134.08N (Friction force)

However, i am really very unsure as to whether i am just supposed to just utilise the horizontal force component of 289.78N to get the acceleration value, as Force = Mass * Acceleration. This is because, there is a reason that the coefficient of sliding friction was provided and therefore, i do not think that it is so simple as to just divide 289.78 over 25kg.

Really tearing my hair out over these questions, as i have been working on these questions for quite a fair bit (3-5 weeks or more ?, for the whole assignment). The worse thing is that my assignment is due on the 25th of May and i am a little worried. Hoping that some one here can simplify and explain these questions to me so that i can understand these biomechanics concepts much better, as i really wanna be well versed in these concepts and not just get these questions over and done with. Thank you.

Regards,

Tze Wei Lim

**Projectile Motion:**1. Neglecting the height of release, a baseball is thrown vertically upwards at 20 m/s, find:

a) vertical velocity after 2 s

b) maximum height

c) total flight time

d) time (seconds) that the projectile is at height 10.5 m.

What would happen to the values obtained in question 1 if the baseball was thrown at an angle of 30 degrees to the ground? (circle one option)

a) Increase, decrease, no change

b) Increase, decrease, no change

c) Increase, decrease, no change

**Basic equations for a strength and conditioning coach:**2. At the instant an athlete is cutting, he is applying 200 N of medio-lateral force and 400 N of anterior-posterior propulsive force. Neglect vertical force in this problem.

a. Based on your knowledge of 3-D axis. Draw the how these two forces would be represented on a diagram, using both direction and magnitude.

b. Calculate the resultant horizontal force of this athlete. Resultant horizontal force = √((400N)2 + (200N)2) = √(160000+40000) = 447.21N (Solved!)

c. Calculate and describe the angle the force is being applied (you may show this on the diagram)

4. Show your work for the following and calculate what happens if an athlete is pulling a sled angle attached to his hip, resulting in an angle of force of pull at 15 degrees to the ground. If the coefficient of sliding friction between the sled and the grass is 0.8 and the sled has a mass of 25 kg, what is the acceleration of the sled if he is constantly pulling with a resultant force of 300 N? (this is a multi-step problem).

So far, i have managed to calculate these steps:

25kg * 9.81 = 245.25N

300sin15° = 77.65N (Vertical component)

245.25N – 77.65N = 167.6N (Reaction force)

300cos15° = 289.78N (Horizontal component)

F = μR, F = 0.8 * 167.6N, F = 134.08N (Friction force)

However, i am really very unsure as to whether i am just supposed to just utilise the horizontal force component of 289.78N to get the acceleration value, as Force = Mass * Acceleration. This is because, there is a reason that the coefficient of sliding friction was provided and therefore, i do not think that it is so simple as to just divide 289.78 over 25kg.

Really tearing my hair out over these questions, as i have been working on these questions for quite a fair bit (3-5 weeks or more ?, for the whole assignment). The worse thing is that my assignment is due on the 25th of May and i am a little worried. Hoping that some one here can simplify and explain these questions to me so that i can understand these biomechanics concepts much better, as i really wanna be well versed in these concepts and not just get these questions over and done with. Thank you.

Regards,

Tze Wei Lim

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