A baseball is thrown from the roof of a 22.0 m building withan initial velocity of magnitude 12.0 m/s and directed at an angleof 53.1

^{0}above the horizontal. (a) What is thespeed of the ball just before it strikes the ground? (b) What isthe answer for part (a) if the initial velocity is53.1^{o }below the horizontal? Use EnergyConservation Principle and ignore air resistance.Comments

## Answers ( 1 )

_{Total,initial}= K_{initial}+P_{initial}^{2}_{o}+ mghThe total energy at the end of the trajectory (at the bottom on theground) is:

_{Total,final}= K_{final}+P_{final}^{2}_{f}+ mg*0= (1/2)mv

^{2}_{f}Since energy isconserved, the amount of it at the beginning is the same as theamount of it at the end. Thus we can set these two values equal toeach other, namely:

^{2}_{o}+ mgh = (1/2)mv^{2}_{f}Noticing that the value of the mass is in all the terms, it can becancelled making the final calculation a little easier,Therefore,

^{2}_{o}+ gh = (1/2)v^{2}_{f}Now solving for vf yields:

_{f}=?(v^{2}_{o}+ 2gh).Plug in your numbers and you will have your answer.

(b) In this case the mass is thrown with the same speed, but at adifferent angle. The thing to notice is that the energy equationdoes not involve any vectors. That is it only involves scalars(numbers with no direction). So since the angle has changed, andthe Conservation of Energy does not depend upon angles, then theanswer will be the same.

NOTE: youmight be wondering "how come?" the reason is that the energy givento the mass in going up in part (a) is lost in dropping from theheight of the trajectory (above the building) to the top of thebuilding. After that theenergy from the top of the building down to the ground is the sameas in part (b), where the energy is calculated from the top to theground directly.

-LD