Biomechanics Assignment: Velocity, Force, and Motion Analysis

This document presents a comprehensive solution to a biomechanics assignment, addressing problems related to velocity, force, and motion analysis. The solution begins with calculating the horizontal and vertical components of a soccer ball's release velocity, considering the angle and initial velocity. It then determines the average velocity of a yacht based on distance and time traveled. The assignment further explores motion using Newton's equations to calculate distance traveled with constant acceleration. Finally, the solution resolves a force vector into its vertical component. The assignment utilizes fundamental physics principles and vector analysis to solve real-world problems, providing a detailed breakdown of each step and referencing relevant scientific literature.

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Biomechanics

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Question-1 : The diagrammatical representation of the problem is given below.
Considering a 2-Dimensional Coordinate System.
v’
Ball i
v x-axis
y-axis
Given : Magnitude of Velocity at which the soccer ball is kicked (|v|) = 23 m/s
According to the Law of Conservation of Momentum, since the ball is initially at
rest, the release velocity will be equal to the velocity at which the ball is kiscked.
So,
v’= v = 23 m/s
Angle between ground and velocity (i) = 18 degrees
Resolving the velocity in horizontal and vertical components-
Horizontal Velocity of Ball (vh’) = 23 x cos (18) = 23 x 0.95 = 21.87 m/s
Vertical Velocity of Ball (vv’) = 23 x sin (18) = 23 x 0.31 = 7.11 m/s
Question-2 :
The Yacht starts at Sydney an after Sydney P Hobart
2.5 hours, it reaches to point P West East
which is 11.6 km from the Sydney.
Total Distance Covered = 11.6 km = 11600m
Total time taken to cover this distance = 2.5 hours = 2.5 x 60 x 60
= 9000 seconds

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Accordng to Spiegel, Lipzchutz & Spellman (2017),
Average velocity of Yacht (vavg) = Total Distance Covered
Total timetaken
= 11600
9000
= 1.29 m/s
Question-3 : u = 0
Let the Maggie and her pram are pushed pushed
from the top of the hill with initial velocity zero . x
Let after covering x distance, the velocity of both
gets to 5.1 m/s v = 5.1 m/s
Since the acceleration is constant i.e.,
a = 3.95 m/s2
Using Newton’s Equation of Motion as stated
in Morin (2009),
v2 – u2 = 2ax
5.12 – 02 = 2 x 3.95 x x
x = 26.01
7.9 = 3.29 m
Question-4 : Consider a 2-Dimensional Coordinate System to represent the foot’s force
in it.
i x - axis
F = 865 N
Here, i = 36 degree y – axis
Vertical Component of Force (Fv) = 895 x sin(i)

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= 895 x sin(36) = 526.07 N = -526.07N
References-
Morin, D. (2009). Introduction to Classical Mechanics. Cambridge : Cambridge
University Press
Speigel, M., Lipschutz, S., Spellman, D. (2017). Vector Analysis.(2nd Ed.) New York :
McGraw - Hill Education