– Horizontal component:
\( v_{0x} = v_p \cos \theta \)
\( v_{0x} = 97.5 \times \cos 50^\circ \approx 97.5 \times 0.6428 \approx 62.73 \, \text{m/s} \)

– Vertical component:
\( v_{0y} = v_p \sin \theta \)
\( v_{0y} = 97.5 \times \sin 50^\circ \approx 97.5 \times 0.7660 \approx 74.72 \, \text{m/s} \)

\( y = y_0 + v_{0y} t – \dfrac{1}{2} g t^2 \)
Since the package lands on the ground, \( y = 0 \) and \( y_0 = h \):
\( 0 = h + v_{0y} t – \dfrac{1}{2} g t^2 \)

\( -\dfrac{1}{2} g t^2 + v_{0y} t + h = 0 \)
Multiply both sides by \(-1\):
\( \dfrac{1}{2} g t^2 – v_{0y} t – h = 0 \)
Write in standard quadratic form \( a t^2 + b t + c = 0 \):
\( a = \dfrac{1}{2} g = 4.9 \)
\( b = -v_{0y} = -74.72 \)
\( c = -h = -732 \)

\( D = b^2 – 4ac \)
\( D = (-74.72)^2 – 4(4.9)(-732) \)
\( D = 5582.78 + 14,361.6 = 19,944.38 \)

\( t = \dfrac{-b \pm \sqrt{D}}{2a} \)
\( t = \dfrac{-(-74.72) \pm \sqrt{19,944.38}}{2 \times 4.9} \)
\( t = \dfrac{74.72 \pm 141.24}{9.8} \)
– Negative root yields negative time (unphysical), so use positive root:
\( t = \dfrac{74.72 + 141.24}{9.8} \approx \dfrac{215.96}{9.8} \approx 22.04 \, \text{s} \)

\( x = v_{0x} t \)
\( x = 62.73 \times 22.04 \approx 1,382.71 \, \text{m} \)

\( v_y = v_{0y} – g t \)
\( v_y = 74.72 – 9.8 \times 22.04 \)
\( v_y = 74.72 – 216 \approx -141.28 \, \text{m/s} \)

\( v = \sqrt{v_{x}^2 + v_{y}^2} \)
\( v = \sqrt{(62.73)^2 + (-141.28)^2} \)
\( v = \sqrt{3,936.57 + 19,948.84} \)
\( v = \sqrt{23,885.41} \approx 154.57 \, \text{m/s} \)

\( \tan \phi = \dfrac{|v_y|}{v_x} = \dfrac{141.28}{62.73} \approx 2.252 \)
\( \phi = \arctan(2.252) \approx 66.66^\circ \)
The angle is below the horizontal since \( v_y \) is downward.