Question

v0x = 7.0 cos(θ) = 7.0 × ( ) = 4.2 m/s
v0y = 7.0 sin(θ) = 7.0 × ( ) = 5.6 m/s

Explanation:

Step1: Analyze \( v_{0x} \) calculation

We know that for a projectile motion (or vector component problem), the horizontal component \( v_{0x} = v_0\cos\theta \). Given \( v_0 = 7.0\space m/s \) and let's assume \( \cos\theta=\frac{4.2}{7.0} = 0.6 \), so \( \theta=\arccos(0.6) \), and \( v_{0x}=7.0\times0.6 = 4.2\space m/s \).

Step2: Analyze \( v_{0y} \) calculation

The vertical component \( v_{0y}=v_0\sin\theta \). Since \( \cos\theta = 0.6 \), we can find \( \sin\theta=\sqrt{1 - 0.6^{2}}=\sqrt{1 - 0.36}=\sqrt{0.64} = 0.8 \), then \( v_{0y}=7.0\times0.8=5.6\space m/s \).

Answer:

The missing values in the calculations are \( \cos\theta = 0.6 \) (for \( v_{0x} \)) and \( \sin\theta=0.8 \) (for \( v_{0y} \)), so the completed calculations are \( v_{0x}=7.0\times0.6 = 4.2\space m/s \) and \( v_{0y}=7.0\times0.8 = 5.6\space m/s \)