Answer:
If a projectile is projected with
velocity u, making an angle \[\theta \] with the horizontal direction, then
Horizontal range, \[R=\frac{{{u}^{2}}}{R}\sin
2\theta \]
And Max. height, \[H=\frac{{{u}^{2}}{{\sin
}^{2}}\theta }{2g}\]
Case (i) if \[\theta
=\alpha \], let \[R={{R}_{1}}\]and \[H={{H}_{1}}\]then \[{{R}_{1}}=\frac{{{u}^{2}}\sin
2\alpha }{g}\] .?. (i)
and\[{{H}_{1}}=\frac{{{u}^{2}}}{2g}{{\sin
}^{2}}\alpha \] ??
(ii)
Case (ii) If \[\theta
=\left( {{90}^{o}}-\alpha \right)\], let \[R={{R}^{2}}\] and \[H={{H}_{2}}\],
then
\[{{R}^{2}}=\frac{{{u}^{2}}\sin
2\left( {{90}^{o}}-\alpha \right)}{g}=\frac{{{u}^{2}}}{g}\sin \left(
{{180}^{o}}-2\alpha \right)=\frac{{{u}^{2}}}{g}\sin 2\alpha \] ??
(iii)
\[{{H}_{2}}=\frac{{{u}^{2}}}{2g}{{\sin
}^{2}}\left( {{90}^{o}}-\alpha \right)=\frac{{{u}^{2}}}{2g}{{\cos }^{2}}\alpha
\] ??
(iv)
From (i) and (iii)
;\[{{R}_{1}}={{R}_{2}}\]
\[{{H}_{1}}+{{H}_{2}}=\frac{{{u}^{2}}}{2g}\left(
{{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)=\frac{{{u}^{2}}}{2g}\]
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