kinetic energy of the stone

Suppose we have a man traveling in an open car (roof open) with speed $v$ towards right (man faces right). He throws a stone (mass $m$) towards right, in his frame-forward with speed $V$.

In the car's frame, the total energy imparted to it was $$E=E_f-E_i=\frac{1}{2}mV^2$$
In the ground frame, the total energy given to it was $$E=E_f-E_i=\frac{1}{2}m((v+V)^2-v^2)=\frac{1}{2}m(V^2+2Vv)$$ What part am I getting wrong on? Is it okay to get such difference?

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Hi Ashish. Don't get tensed that I tagged this as homework. This is a nice example of what our homework tag suggests. It is misused often. But, these questions still apply to our homework tag ;-) – Waffle's Crazy Peanut Mar 23 '13 at 10:43
it was just to inform, i didn't mind anyway – Ashish Gaurav Mar 23 '13 at 10:44

Suppose you throw two rocks with the same velocity $V$. The first one, you throw at a target moving alongside the car. The second one you throw at a stagnant target by the side of the road. Which one will take more damage?
Brain teaser: Compare a car crashing at a parked car with velocity $V$ to a heads-on collision between to cars driving at velocity $\frac{1}{2}V$. Is this the same collision? (Extensive answers can be found elsewhere on this site)
i think the stagnant target would be affected more, because then the impact would be more. If the car was with velocity $v$ and the stone had it as $q$ then the change in momentum would be from $m(v+q)$ to $0$. In the other case, only the throwing velocity causes the momentum change. Is this correct? – Ashish Gaurav Mar 23 '13 at 10:50