ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground…

ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. what is the approximate horizontal component of the initial velocity?\n0.64 m/s\n0.77 m/s\n10.3 m/s\n12.3 m/s

ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. what is the approximate horizontal component of the initial velocity?\n0.64 m/s\n0.77 m/s\n10.3 m/s\n12.3 m/s

Answer

Explanation:

Step1: Recall velocity - component formula

The formula for the horizontal component of velocity $v_x$ is $v_x = v\cos\theta$, where $v$ is the initial velocity and $\theta$ is the angle of projection.

Step2: Substitute given values

Given $v = 16$ m/s and $\theta=40^{\circ}$. So $v_x=16\times\cos40^{\circ}$. We know that $\cos40^{\circ}\approx0.766$. Then $v_x = 16\times0.766 = 12.256\approx12.3$ m/s.

Answer:

D. 12.3 m/s