A boy throws a stone straight upward with an initial velocity of 5 m/s from the surface of the Earth. What is the maximum height above the starting point will the stone reach before it starts falling back?
"h=\\frac{v^2-v_0^2}{2g}=\\frac{0-5^2}{2\\cdot(-9.8)}=1.28\\ (m)" . Answer
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