Is there a relationship between acceleration and jump height

AI Thread Summary
The discussion explores the relationship between head acceleration and jump height, particularly for a 5-foot-2-inch boy weighing 120 pounds. Participants suggest that body acceleration into the floor before jumping is crucial, as it may influence the duration of the jump in the air. There is confusion regarding the definitions of head and body acceleration, prompting a request for clarification on these concepts. Additionally, the conversation touches on the forces affecting acceleration and their connection to jump height. Understanding these dynamics may help clarify the physics behind jump performance.
Tony H
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Homework Statement


Is there a relationship between head acceleration and jump height say for a 5 foot 2 inches boy weighing 120 lb?

Homework Equations

The Attempt at a Solution


I think there is a relationship between body acceleration into the floor before jumping, but wondering if there is one for head acceleration. Faster body acceleration getting off the floor into the jump means you will be in the air longer, I believe.

Any help would be appreciated, thank you
 
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I don't follow the question. Can you explain a little more clearly. What is head acceleration? body acceleration? faster body acceleration?
 
Think of it this way: What are the forces affecting acceleration? And what are they liked to?
 
Tony H said:
I think there is a relationship between body acceleration into the floor before jumping, but wondering if there is one for head acceleration. Faster body acceleration getting off the floor into the jump means you will be in the air longer, I believe.

Any help would be appreciated, thank you
What conservation law can you connect with the height of the jump?
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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