What is the kenetic energy of 1-kilogram ball as thrown into the air with an initial velocity of 30 m/sec?

This problem is one tht directly involve gravity, kinetic energy is 1/2 massXvelocity^2, so .5*1*30^2=450joules, the unit for energy. 

The potential energy of it is mass times height times acceleration to gravity, 9.81. This is trickier as you have to calculate height first. 

To do this, you use the formula velocity final squared= velocity initial squared + 2*a*x, a=-9.81, final velocity = 0, initial=30, so 0=900-2*9.81x, 900/19.62=45.87ft. 

Height=45.87ft, PE=m*a*h, PE=1*9.81*45.87, so Potential energy is 450 joules. This makes sense since with energy conserved, when KE=0, PE must equal the previous KE, and when PE=0, KE must equal that constant 

Formulas: 

KE=1/2*m*v^2 
PE=m*g*h 
Vf^2=Vi^2+2ax 
Vf=0, Vi=30, g=9.81, m=1

    5.5 n======    assuming that      we have      m = 0.145 kg    a = 38 m/s²therefore    5.51 n, but two significant figures.final answer: 5.5 n