How Do You Calculate the Height of a Rock at Half Its Initial Speed?

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To calculate the height of a rock at half its initial speed, one can use the conservation of energy principle, where the kinetic energy (KE) lost equals the potential energy (PE) gained. The rock, with a mass of 22.8 kg and an initial KE of 28 Joules, is thrown upwards at an initial velocity of 1.56721 m/s. At half its initial speed, the rock's velocity is 0.783605 m/s, and the corresponding height can be determined by equating the lost KE to the gained PE. Alternatively, kinematic equations can also be applied, specifically v_f^2 = v_i^2 + 2ad, to find the height. The conservation of energy method was confirmed as effective for this calculation.
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ok you have a rock of mass 22.8 KG
It has an initial KE of 28 Joules
This rock is thrown upwards with an initial velocity of 1.56721 m/s
It peak height is .122807 ( i hope that's right :confused:)
What is the height of the rock at half its initial speed?
If someone could just give me the right formula or head me in the right direction that would be great. :D
 
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One way is to use conservation of energy. As it rises, it loses KE (figure out how much) and gains PE.

Another way is to use kinematics: v_f^2 = v_i^2 + 2ad
 
Thankyou, got it with conservation of energy. :D
 
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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