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question 8 of 25
question 8
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a football is kicked with a speed of 18 m/s at an angle of 65° to the horizontal. what are the respective horizontal and vertical components of the initial velocity of the football?
○ (v_{ox}=7.6 m/s, v_{oy}=16 m/s)
○ (v_{ox}=16 m/s, v_{oy}=7.6 m/s)
○ (v_{ox}=13 m/s, v_{oy}=8.4 m/s)
○ (v_{ox}=9 m/s, v_{oy}=9 m/s)
○ (v_{ox}=8.4 m/s, v_{oy}=13 m/s)

Explanation:

Step1: Calculate horizontal component

The formula for the horizontal - component of initial velocity is $v_{ox}=v_0\cos\theta$. Given $v_0 = 18\ m/s$ and $\theta = 65^{\circ}$, then $v_{ox}=18\cos65^{\circ}\approx18\times0.4226\approx7.6\ m/s$.

Step2: Calculate vertical component

The formula for the vertical - component of initial velocity is $v_{oy}=v_0\sin\theta$. Given $v_0 = 18\ m/s$ and $\theta = 65^{\circ}$, then $v_{oy}=18\sin65^{\circ}\approx18\times0.9063\approx16\ m/s$.

Answer:

$v_{ox}=7.6\ m/s, v_{oy}=16\ m/s$