Answer:
Let the projectile go from O to P in time \[{{\text{t}}_{\text{1}}}\left(
=\text{1s} \right)\]and from O to Q in time \[{{\text{t}}_{\text{2}}}=\left(
\text{3s} \right)\]
Initial vertical velocity of particle at \[\text{O}=\text{ucos}\theta
\].
Taking vertical upward motion of particle from O to P and O to Q,
we have
\[\text{h}=\text{ucos}\theta
{{\text{t}}_{\text{1}}}-\frac{1}{2}gt_{1}^{2}=\text{ucos}\theta
{{\text{t}}_{\text{2}}}\frac{1}{2}gt_{2}^{2}\]
or \[\text{u cos}\theta \times
\text{1}\frac{1}{2}\text{g}\times {{\text{1}}^{\text{2}}}=\text{ucos}\theta
\times \text{3}~\frac{1}{2}\text{g}{{\text{3}}^{\text{2}}}\]
or \[\text{u cos}\theta
(\text{3}-\text{1})=~\frac{g}{2}\left( \text{9}\text{1}
\right)=~\frac{9.8}{2}\times \text{8}=\text{4}.\text{9}\times \text{8}\]
or \[\text{u cos}\theta =\text{
}\frac{\text{4}.\text{9}\times \text{8}}{2}\times \text{4}=\text{19}.\text{6
m}/\text{s}\]
Max. height, \[\text{H}=\frac{{{u}^{2}}{{\cos }^{2}}\theta
}{2g}=\frac{{{(19.6)}^{2}}}{2\times 9.6}=19.6m\]
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