What is the net force's direction of an object at its highest point?

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At an object's highest point, the net force is directed downwards due to the constant presence of gravity. Gravity acts on the object regardless of its motion, ensuring a downward pull. This understanding is crucial for analyzing forces in physics. The confirmation of this concept highlights the importance of gravitational force in determining net force direction. Overall, gravity remains the dominant force acting on the object at its peak.
Mikasun1108
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Homework Statement
A ball is thrown upwards towards the sky. What is the direction of the net force when it reaches its highest point?
a. Left
b. Right
c. No direction
d. Upwards
e. Downwards
Relevant Equations
Regarding free fall
I'm not sure of my answer but it is e(downwards). I arrive at this answer since gravity always exists so no matter what happens it always pulls an object downwards. But I am not sure. Thank you
-sun1108
 
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Right.
 
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haruspex said:
Right.
Thank you for your help haruspex. Really appreciate it! hope you have a great day! :)
 
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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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