A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is $$10\,m/s,$$  then the maximum height attained by the stone is $$\left( {g = 10\,m/{s^2}} \right)$$

Solution :
Let $$u$$ be the initial velocity and $$H$$ be the maximum height attained.
$$\eqalign{ & {\text{At height }}h = \frac{H}{2},\,{\text{we have }}v = {v_1}\, = 10\,m/s \cr & {\text{From third equation of motion, }}v_1^2 = {u^2} - 2gh\left( {{\text{Negative sign indicates that velocity and acceleration are in opposite direction}}} \right) \cr & {\text{or}}\,{\left( {10} \right)^2} = {u^2} - 2g\frac{H}{2}\,......\left( {\text{i}} \right) \cr & {\text{At}}\,{\text{height}}\,H,\,{v_2} = 0 \cr & v_2^2 = {u^2} - 2gH\,\,{\text{or}}\,\,0 = {u^2} - 2gH\,......\left( {{\text{ii}}} \right) \cr & {\text{Subtract Eq}}{\text{. }}\left( {{\text{ii}}} \right){\text{ from Eq}}{\text{. }}\left( {\text{i}} \right),{\text{ we get}} \cr & {\left( {10} \right)^2} = 2g\frac{H}{2}\,\,{\text{or}}\,\,H = \frac{{{{\left( {10} \right)}^2}}}{g} \cr & {\text{or}}\,H = \frac{{{{\left( {10} \right)}^2}}}{{10}} = 10\,m \cr} $$
Alternative
Maximum height attained by the stone
$$\eqalign{ & H = \frac{{{u^2}}}{{2g}} \cr & {\text{When,}}\,H = \frac{H}{2},u = 10\,m/s \cr & \frac{H}{2} = \frac{{{{\left( {10} \right)}^2}}}{{2g}}\,\,{\text{or}}\,\,H = \frac{{100}}{{10}} = 10\,m \cr} $$