Answer:
If \[\theta \] is the angle of projection, then
velocity of projectile at height point \[~=\text{ucos}\theta \]
As per question, \[\text{u cos}\theta
=\frac{50}{100}u=\frac{1}{2}u\] or \[\text{cos}\theta =\frac{1}{2}=\text{cos 6}0{}^\circ
\]
or \[\theta =\text{6}0{}^\circ \]
Horizontal range, \[\text{R}=\frac{{{u}^{2}}\sin 2\theta
}{g}=\frac{{{u}^{2}}}{g}\text{sin 2}\times \text{6}0{}^\circ
=\frac{{{u}^{2}}}{g}\times \sin \left( {{180}^{o}}-{{60}^{o}} \right)\]
\[=\frac{{{u}^{2}}}{g}\sin
{{60}^{o}}=\frac{{{u}^{2}}}{g}\times \frac{\sqrt{3}}{2}\]
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