How Do You Calculate the Maximum Height of an Object Thrown Upward?

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To calculate the maximum height of an object thrown upward with an initial speed of 21 m/s, the problem can be reframed to analyze its motion at different points. At two-thirds of the maximum height, the object's speed is 21 m/s, and at the maximum height, the speed is 0 m/s. Using the kinematic equation v_f^2 = v_i^2 + 2ad, where the distance traveled from two-thirds to maximum height is h/3, allows for solving the maximum height. The discussion emphasizes the importance of breaking down the problem into manageable segments to apply the kinematic equations effectively. This method provides a clear path to determining the maximum height of the object.
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An object is thrown vertically upward such that it has a speed of 21 m/s when it reaches two thirds of its maximum height above the launch point. Determine this maximum height.

I can't figure what to do since initial velocity isn't given and I don't see a way to solve for it.
 
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Reframe the problem this way: An object is thrown upward with an initial speed of 21 m/s and rises to a height of h/3. Solve for h.
 
Oh I like that suggestion Doc Al. Very clever.

I would use a kinematic equation to help get your answer... this one perhaps?

v_{f}^2=v_{i}^2+2ad
 
That equation doesn't really help me though, Jameson. It has two variables in it that I don't know the value of.

And I kind of see where you're going, Doc Al, but why h/3? It seems like 2h/3 would be what I'd want to solve for...
 
What doc al is saying is you can find the maximum heigth because at the point \frac{2}{3}h the initial velocity would be 21ms/2 and at maximum height (which is \frac{1}{3}h above that point) you know the ball would come to rest. This would allow you to use Jameson's equation to calculate the maximum height.
 
finlejb said:
And I kind of see where you're going, Doc Al, but why h/3? It seems like 2h/3 would be what I'd want to solve for...
Hootenanny already explained it, but let me put it this way. You have three points of interest:
(1) start: y = 0; v = ?
(2) "mid" point: y = 2h/3; v = 21 m/s
(3) top: y = h; v = 0 m/s

What I'm suggesting is that between points 2 & 3 the object travels a distance of h/3. A perfect opportunity to apply Jameson's kinematic equation.
 
Thread 'Variable mass system : water sprayed into a moving container'

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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