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

AI Thread Summary
The discussion centers on calculating the take-off speed and distance for a high jumper whose center of gravity is 1.1m above the ground while clearing a height of 1.8m. The significance of the center of gravity is clarified, indicating it should be subtracted from the jump height for calculations. The initial take-off velocity is determined to be approximately 4.279 m/s, with a required distance of 0.808m from the bar. Additionally, there is a query about using angles for downward projections, emphasizing the importance of correctly applying trigonometric functions to resolve vectors into their x- and y-components. Overall, the calculations and concepts discussed are deemed reasonable and correct.
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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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