Question

rome kicks a long - range ball with an initial velocity of 18 m/s at an angle of 40° relative to the ground. what is the approximate horizontal component of the initial velocity?
0 2.64 m/s
0 9.27 m/s
0 10.3 m/s
0 13.8 m/s

Explanation:

Step1: Recall the formula for horizontal - component of velocity

The formula for the horizontal component of the initial velocity $v_{0x}$ is $v_{0x}=v_0\cos\theta$, where $v_0$ is the initial velocity and $\theta$ is the angle of projection with respect to the ground.

Step2: Substitute the given values

We are given that $v_0 = 18\ m/s$ and $\theta = 42^{\circ}$. So, $v_{0x}=18\times\cos(42^{\circ})$.
We know that $\cos(42^{\circ})\approx0.743$. Then $v_{0x}=18\times0.743 = 13.374\approx13.4\ m/s$.

Answer:

The closest option is 13.3 m/s.