b
(b) Let particle thrown with velocity \(u\) and its maximum height is \(H\) then \(H = \frac{{{u^2}}}{{2g}}\)

When particle is at a height \(H/2\), then its speed is \(10\, m/s\)

From equation \({v^2} = {u^2} - 2gh\)

\({(10)^2} = {u^2} - 2g\left( {\frac{H}{2}} \right) = {u^2} - 2g\frac{{{u^2}}}{{4g}}\)\( \Rightarrow {u^2} = 200\)

Maximum height \( \Rightarrow H = \frac{{{u^2}}}{{2g}} = \frac{{200}}{{2 \times 10}} = 10\;m\)