Question

10 a soccer ball is kicked with an initial velocity of 18.0 meters per second, at an angle of 24.0° above horizontal. what is the horizontal component of the velocity vector when the ball is landing on the ground again? 16.4 m/s 7.32 m/s -16.4 m/s 7.32 m/s

Explanation:

Step1: Recall the formula for horizontal component of velocity

The formula for the horizontal component of velocity \(v_x\) is \(v_x = v_0\cos\theta\), where \(v_0\) is the initial velocity and \(\theta\) is the angle of projection.

Step2: Substitute the given values

Given \(v_0 = 18.0\space m/s\) and \(\theta = 24.0^{\circ}\).

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In projectile motion, the horizontal component of velocity remains constant throughout the motion (neglecting air - resistance). So when the ball lands, the horizontal component of velocity is the same as the initial horizontal component of velocity.

Answer:

16.4 m/s