Question

6 a soccer ball is kicked with an initial velocity of 14.1 meters per second at an angle of 31.6° above horizontal. what is the horizontal component of the velocity vector when the ball is landing on the ground again?
12.0 m/s
7.38 m/s
-0.81 m/s
-7.38 m/s

Explanation:

Step1: Recall the formula for horizontal component of velocity

The horizontal component of velocity \(v_x = v_0\cos\theta\), where \(v_0\) is the initial velocity and \(\theta\) is the angle of projection. There is no acceleration in the horizontal direction (assuming no air - resistance), so the horizontal component of velocity remains constant throughout the motion.

Step2: Substitute the given values

Given \(v_0=14.1\ m/s\) and \(\theta = 31.6^{\circ}\).

$$v_x=14.1\times\cos(31.6^{\circ})$$

We know that \(\cos(31.6^{\circ})\approx0.851\)

$$v_x = 14.1\times0.851\approx12.0\ m/s$$

Answer:

A. \(12.0\ m/s\)