Question

7 a ball is launched with an initial velocity of 44.1 meters per second, at an angle of 34.4° above horizontal. at the peak of its trajectory, what are the components of the balls velocity vector?

Explanation:

Step1: Analyze horizontal velocity component

In projectile motion, the horizontal component of velocity \(v_x = v_0\cos\theta\). Given \(v_0 = 44.1\ m/s\) and \(\theta=34.4^{\circ}\), then \(v_x=44.1\times\cos(34.4^{\circ})\).
Using a calculator, \(\cos(34.4^{\circ})\approx0.825\), so \(v_x = 44.1\times0.825\approx36.4\ m/s\).

Step2: Analyze vertical velocity component

At the peak of the trajectory, the vertical component of velocity \(v_y = 0\ m/s\) (because the ball stops moving upward at the peak and the vertical - direction velocity changes from positive (upward) to negative (downward) at that point).

Answer:

\(\overrightarrow{v_x}=36.4\frac{m}{s},\overrightarrow{v_y} = 0\frac{m}{s}\) (the second option)