Projectile Motion: High Jumper Take Off Velocity & Distance from Bar

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Homework Help Overview

The discussion revolves around a high jumper's take-off velocity and the distance from the bar during a jump, specifically analyzing the significance of the jumper's center of gravity in relation to the height cleared. The subject area includes concepts from kinematics and projectile motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of the center of gravity on the jump, with one questioning whether to subtract the center of gravity height from the jump height for calculations. Others discuss the calculations related to the initial velocity and time to reach the highest point, while also raising questions about the correct interpretation of angles in projectile motion.

Discussion Status

The discussion is active, with participants providing feedback on calculations and addressing conceptual questions. Some guidance has been offered regarding the use of trigonometric functions and the importance of resolving vectors into components.

Contextual Notes

Participants are navigating the complexities of projectile motion and the specific parameters of the problem, including the height of the center of gravity and the angles involved in the calculations. There is an emphasis on understanding the physical setup before proceeding with calculations.

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Homework Statement


A high jumper whose center of gravity is 1.1m above the ground h been clearing 1.8m using a western roll in which the take off velocity is at an angle 60deg with the horizontal. With what speed must be take off? how far back from the bar must he take off?

I understand the question except for the 1.1m above ground center of gravity.
What significance does this have?


Homework Equations


kinematics


The Attempt at a Solution


I can't proceed until I understand what the center of gravity is about.
 
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His centre of gravity must move up and over the bar to make the jump
 
Would it be alright if i subtracted 1.1m from 1.8m and then began my calculations?
 
Yes, that's fine
 
calculations
Vfy^2=Voy^2+2(-9.81)S
0=(VoSin60)^2 + 2(-9.81)(0.7)
Vo = 4.279m/s -- is the initial velocity he must take off with.

0.7 = -0.5(-9.81)t^2
t= 0.3778secs -- is the time it takes to reach the highest point i.e. the bar

x=voxt
x=4.279cos60(0.3778)
x=0.808m -- the distance away from the bar he must jump

looks reasonable, can someone concur?

another question i have is, if an object is directly projected downwards at an angle with the horizontal, which is the angle of depression.

Do i use this angle for Vox and Voy or 90 minus that angle?
 
Last edited:
is that correct or do i have to minus by 90?
 
Your calculations look correct to me.

another question i have is, if an object is directly projected downwards at an angle with the horizontal, which is the angle of depression.

Do i use this angle for Vox and Voy or 90 minus that angle?

You shouldn't liberally apply cosines and sines without thinking about what you're doing. Remember, the whole point of using the trig functions is to resolve a vector into mutually perpendicular axes, in this case, to resolve the velocity into the x- and y-directions. Draw a diagram with the velocity and the angle, then axes representing the x- and y-axes, and see what trig functions you need to resolve it into a given direction.
 

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