a.) Determine the initial horizontal and vertical velocity.
\[\cos{30^{\circ}}= \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{x}{300}\]
\[\implies x = 300 \times \cos{30^{\circ}}\]
\[\therefore x= 259.81m/s\]
b.) Determine the time to reach maximum height and the total time of the flight.
Total Time of Flight \((t)=\frac{2 u \sin \theta}{g}\)
\[t=\dfrac{2\cdot 300 \cdot \sin 30^{\circ}}{10} = 30s\]
c.) Determine the maximum height.
The maximum height of the projectile is given by the formula:
\[
H=\dfrac{v_{0}^{2} \sin ^{2} \theta}{2 g}
\]
Assuming \(g=10m/s^2\)
\[
H=\dfrac{300^{2} \sin ^{2} 30^{\circ}}{2\cdot 10}
\]
\[
H=\dfrac{300^{2} \sin ^{2} 30^{\circ}}{2\cdot 10} = 1125m
\]
d.) Determine the horizontal range.
Horizontal Range \((R)=\dfrac{u^{2} \sin 2 \theta}{g}\)
Assuming \(g=10m/s^2\)
\[(R)=\dfrac{300^{2} \sin{60^{\circ}}}{10} = 7794.23m\]