If a 2 kg balls thrown straight upward with a KE of 500 j. what maximum height will it reach?

I think its 25.51 m!!

Explanation:

Your answer is, 25.51m

Solution to the problem

KEi +PEi = KEf + PEf

KEi = 500, PEi = 0, KEf = 0, PEf = 500

PE = fg * m * h

500 = 9.8 * 2 * h

h = 25.51m

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol  

b. a hockey puck sliding across the ice  

c. a compressed spring  

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice   is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice   is the correct option.