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:
[tex]E_k = 0.5 mv_1^2[/tex]
[tex]E_p = mgh[/tex]
And as it hits the ground:
E_k = 0.5mv_2^2[/tex]
[tex]E_p = 0
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).
The initial velocity, Viy
The final displacement, Dy
The acceleration, a
You are looking for the final velocity, Vfy
Separate names with a comma.