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A stone of mass m is thrown straight up into the air with speed v_o. While the flight it feels a force of air resistance of magnitude f. Determine the speed of the stone when it hits the ground. You will need to combine energy conservation equations for the upward and downward parts of the motion.
This is what I did...
upward
(1/2)m(v_o)^2 - F = (1/2)(m)(v_f)^2 + mgh
downward
mgh - F = (1/2)m(v_o)^2
Am I correct so far? How do I combine them?
The correct answer is v = (v_o) SQRT(mg-f/mg+f). How do I get there?
This is what I did...
upward
(1/2)m(v_o)^2 - F = (1/2)(m)(v_f)^2 + mgh
downward
mgh - F = (1/2)m(v_o)^2
Am I correct so far? How do I combine them?
The correct answer is v = (v_o) SQRT(mg-f/mg+f). How do I get there?