Question

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?
0.64 m/s
0.77 m/s
10.3 m/s
12.3 m/s

Explanation:

Step1: Recall velocity - component formula

The horizontal - component of the initial velocity $v_x$ is given by $v_x = v_0\cos\theta$, where $v_0$ is the initial velocity and $\theta$ is the angle of projection.

Step2: Substitute given values

We are given that $v_0 = 16$ m/s and $\theta=40^{\circ}$. So, $v_x = 16\times\cos(40^{\circ})$.
Since $\cos(40^{\circ})\approx0.766$, then $v_x = 16\times0.766 = 12.256\approx12.3$ m/s.

Answer:

12.3 m/s