Answer:
(a, b) \[H\,=\frac{{{u}^{2}}{{\sin
}^{2}}\theta }{2g}\]
Since \[{{H}_{1}}>{{H}_{2}}\]
Also \[T\propto \,\sin
\,\theta \,\,\therefore \,\,{{T}_{1}}>\,{{T}_{2}}.\]
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