Please help with this physics problem! thanks so much!

Question

A 2-kg stone is dropped from a height h and strikes the ground with a velocity of 24 m/s. (a) Find the kinetic energy of the stone as it strikes the ground and the height h from which it was dropped. (b) Solve part a, assuming that the same stone is dropped on the moon. (acceleration of the gravity on the moon=1.62 m/s^2). please help! im not sure if im doing it right! thank you so much!

Answer 1

a) use formula for kinetic enegy

K.E.=0.5mv^2

= 0.5*2*24^2

= 576 J

for the height use law of conservation of energy to calculate height assuming no air resistance; energy at the end must be the same as energy at the beginning, using this fact we already know the potential energy (velocity is zero at beginning) which can allow us to get height

P.E.= mgh

576= 2*9.8*h

h= 29.3877551 metres

b) kinetic energy would still be the same , only gravitational acceleration would be different and since it is less than that means it must be dropped at higher height to get same amount of energy as on Earth

P.E.= mgh

576=2*1.62*h

h= 177.77777... metres

hope this helps

Answer 2

m = 2 kg

v = 24 m/s

a) KE = 1/2 mv^2  = (1/2)(2)(24)^2

KE = 576 N-m

using kinematic equation

Vf^2 - Vo^2 = 2gh

24^2 = 2(9.8)h

576/(2*9.8) = h

h = 29.4 m

using energy conservation law

KE = PE = 576 = mgh

h = 576/(2*9.8)

h = 29.4 m

Answer 3

Post how much you have already done even if you are not sure of its accuracy. Then, I will make correction if needed.

Answer 4

The whole point of homework is to do it yourself and be prepared for test, I  don't think someone should tell him the answer, maybe explain to him the problem.