Question

A. what is the kinetic energy of a 1 kg ball thrown in the air with an initial velocity of 30 m/s

B. How high did the ball travel? KE=PE

Answer 1

By Definition of Kinetic Energy, it is valid:


`E_c= (mv^2)/(2) \\ E_c= (1* 30^2)/(2) \\ \boxed {E_c=450~J}`

Using the Torricelli's Equation in vertical direction on the maximum height, remember that the final velocity at this point is zero, we have:


`v^2=v_(o)^2+2a\Delta S \\ 0=30^2+2(-10)H \\ H= (900)/(20) \\ \boxed {H=45~m}`

If you notice any mistake in my english, please let me know, because my english level is still low.