2D kinematics -- Calculate the acceleration of the jumping athlete

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The discussion revolves around calculating the acceleration of a jumping athlete, initially leading to confusion over an answer of 11 m/s². The participant acknowledges making an error and seeks assistance in determining the direction of acceleration, specifically mentioning a "backward 55 degrees up" reactive force. Ultimately, they express satisfaction in resolving the initial calculation issue but still need to clarify the direction of acceleration. The conversation highlights the importance of considering all forces in kinematic calculations. Accurate direction determination is essential for complete understanding of the athlete's motion.
Stewkatt
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Homework Statement
An athlete with a mass of 62 kg jumps and lands on his feet. The ground exerts a total force of 1.1 x 10^3 N [backward 55 degrees up] on his feet. Calculate the acceleration of the athlete
Relevant Equations
F=delta p/delta t =ma
DCD301E4-C8E1-43CD-A362-D64E05F31F14.jpeg

this is my work but the answers say 11 m/s^2 so I made an error somewhere. Also if someone could help me with solving the direction for the acceleration, that would be greatly appreciated.
 
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Perhaps because you have omitted the "backward 55 degrees up" direction of the reactive force.
 
image.jpg

yay, I figured it out. I still have to find the direction of acceleration and that’s it.
 
Stewkatt said:
yay, I figured it out. I still have to find the direction of acceleration and that’s it.
Did you find the direction of acceleration?
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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