d
\(\begin{array}{l}
Let'u'\,be\,the\,velocity\,\\
\therefore u = 48m/s,\,Given,\,g = 32\\
At\,{\rm{maximum}}\,height\,v = 0\\
Now,\,we\,know\,{v^2} = {u^2} - 2gh\\
\Rightarrow 0 = {\left( {48} \right)^2} - 2\left( {32} \right)h \Rightarrow h = 36\\
Maximum\,height\, = 36 + 64 = 100\,mt
\end{array}\)