Question

1. a paper airplane is thrown with an initial velocity of 2.4 m/s and an angle of 60°. find the ( v_{ox} ).

2.08 m/s
1.70 m/s
1.2 m/s
0.62 m/s

Explanation:

Step1: Recall the formula for horizontal component of velocity

The formula for the horizontal component of velocity \(v_{0x}\) is \(v_{0x}=v_0\cos\theta\), where \(v_0\) is the initial velocity and \(\theta\) is the angle of projection.

Step2: Substitute the given values

Given \(v_0 = 2.4\space m/s\) and \(\theta = 60^{\circ}\), and \(\cos60^{\circ}=\frac{1}{2}\).
So, \(v_{0x}=2.4\times\cos60^{\circ}\)

Step3: Calculate the value

\(v_{0x}=2.4\times\frac{1}{2}=1.2\space m/s\)

Answer:

C. 1.2 m/s