Help please, calculation involving height, speed and gravity

AI Thread Summary
The discussion explains how to solve a physics problem involving height, speed, and gravity without needing the mass of the ball, as it cancels out during calculations. It emphasizes the law of conservation of energy, stating that the total kinetic and potential energy at the start equals that at the moment the ball hits the ground. The equations for kinetic and potential energy are provided, demonstrating that mass appears in all terms and can be eliminated. While it's acceptable to omit mass for simplification, it's advised to avoid canceling terms prematurely in calculations. The thread concludes by referencing kinematic equations to find the final velocity.
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Solved. Thanks.
 
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You can actually do the problem without knowing the mass of the ball. Instead of using 155g, do all of the calculations using 'm', and you should find that it will cancel out in the process of the calculation.

Consider the law of conservation of energy -- the sum of the kinetic and potential energies at the instant the ball is thrown should be equal to the sum of energies at the instant the ball hits the ground.
 
So I can use the two above equations but not include the 'm'? Thanks for your help.
 
Well you would still include the m, but they will eventually cancel out, so omitting the m right from the start will not make a difference (although it is probably a bad habit to cancel things out before it actually happens during the process of the calculation).

What i mean is that initially, as the ball is released, it has kinetic and potential energy (with reference to the ground) given by:

E_k = 0.5 mv_1^2
E_p = mgh

And as it hits the ground:

<br /> E_k = 0.5mv_2^2
E_p = 0<br />

Once you create an equation from these terms, you will see that every term has an 'm' in it which can be eliminated by dividing everything by m.
 
Consult the Kinematic Equations (see attached).

You know:
The initial velocity, Viy
The final displacement, Dy
The acceleration, a

You are looking for the final velocity, Vfy
 

Attachments

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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