A 12.6 kg rock is sitting at rest at a height of h1 = 10.7 m above the ground. What is the kinetic energy, in J, of the ball when it has dropped to a height of h2 = 4.6 m above the ground?

19 hertz bruh u trippin? ?

Pe = 1/2k(x)^2  and .05n/m = .05x10^-9m = x 

Ke reqd to move from 0 to 50 will be proportional to 50x50 or 2500. this is equal to w.  to change vel from 50 to 150 you need energy prop to 22500 - 2500 = 20,000. this corresponds to 8w.  best answer is a) . work reqd is prop to velocity squared.