Answer
a). $v_{y}>v_{x}$ i.e. option (1) is correct.
b). 31.32 mi.
Work Step by Step
a). The orientation angle is $20^{\circ}$ with the vertical.
So, the horizontal component of the velocity is smaller, and horizontal component is larger.
So, $v_{y}>v_{x}$ i.e. option (1) is correct.
b). The horizontal component = $v_{x}=2000sin20^{\circ}=684 mi/hr$
And the vertical component =$v_{y}=2000cos20^{\circ}=1879 mi/hr$
Hence part (a) is verified.
Also, During this time the missile will rise =$1879\times \frac{1}{60}\times1 \,mi=31.32 mi$
