Dropping the Ball: Solving a Physics Problem With 6.8 Joules of Force

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To determine the height from which to drop a 0.535 kg ball to achieve an impact force of 6.8 Joules, one must apply the principles of energy conservation. The kinetic energy (KE) at the moment of impact should equal the potential energy (PE) at the drop height, leading to the equation KE = PE. The force experienced upon impact is related to the change in momentum, which is influenced by how quickly the ball stops upon hitting the ground. By equating the desired kinetic energy of 6.8 Joules to the potential energy formula (PE = mgh), where 'm' is mass, 'g' is acceleration due to gravity, and 'h' is height, the appropriate height can be calculated. This approach effectively links the concepts of force, momentum, and energy in solving the problem.
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So, here is the problem:

I have a .535kg ball that I drop from an unkown height, to create an impact with a force of 6.8 Joules.

I need to figure out what height to drop the ball from.

I figure it has something to do with accleration due to gravity and Netwons law F=mA or possibly a kinetic energy equation.

Right now I am scrambling to find the answer.

Can anyone please help?
 
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You need to define mathematically what 'impact' is. If you drop something it acquires momentum. A force is exerted when there's a change in the momentum, like when the dropped thing hits the ground. The force depends on how quickly the object stops. The faster it stops the greater the force.
 
Well, if you define impact as energy, which is suggested by your units, then what you need to do is apply the Conservation of Energy. In your case, you wan't a KE of 6.8J, hence you can equtae it with potential energy.
 
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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