Question

1. a ball is launched with an initial velocity of 27.7 meters per second at an angle of 12.5° above horizontal. at the peak of its trajectory, what are the components of the balls velocity vector?

\\( \vec { v _ { x } } = 6.00 \mathrm { m } / \mathrm { s }, \vec { v _ { y } } = 27.0 \mathrm { m } / \mathrm { s } \\)
\\( \vec { v _ { x } } = - 6.00 \mathrm { m } / \mathrm { s }, \vec { v _ { y } } = 0 \mathrm { m } / \mathrm { s } \\)
\\( \vec { v _ { x } } = 27.7 \mathrm { m } / \mathrm { s }, \vec { v _ { y } } = 6.00 \mathrm { m } / \mathrm { s } \\)
\\( \vec { v _ { x } } = 27.0 \mathrm { m } / \mathrm { s }, \vec { v _ { y } } = 0 \mathrm { m } / \mathrm { s } \\)

Explanation:

Step1: Analyze horizontal velocity component

In projectile motion, the horizontal component of velocity \(v_x = v_0\cos\theta\), where \(v_0 = 27.7\space m/s\) and \(\theta=12.5^{\circ}\). There is no acceleration in the horizontal direction (\(a_x = 0\)), so the horizontal velocity remains constant throughout the motion.

$$v_x=27.7\cos(12.5^{\circ})$$

$$v_x\approx27.7\times0.9763\approx27.0\space m/s$$

Step2: Analyze vertical velocity component

At the peak of the trajectory, the vertical component of velocity \(v_y = 0\space m/s\). This is because the ball stops moving upward (the vertical - motion is under the influence of gravity \(a_y=-g\), and at the maximum height, the instantaneous vertical velocity is zero).

Answer:

\(\vec{v}_x = 27.0\space m/s,\vec{v}_y = 0\space m/s\) (the fourth option)