Question

6 a ball is launched with an initial velocity of 44.1 meters per second, at an angle of 34.4 above horizontal. when the ball reaches the ground again, what is its velocity vector?

Explanation:

Step1: Calculate the horizontal component of initial velocity

The horizontal component of velocity \(v_{x}\) remains constant throughout the projectile motion. The formula for the horizontal component of initial velocity is \(v_{x}=v_{0}\cos\theta\), where \(v_{0} = 44.1\ m/s\) and \(\theta=34.4^{\circ}\).

$$v_{x}=44.1\times\cos(34.4^{\circ})$$

$$v_{x}=44.1\times0.825$$

$$v_{x}\approx36.4\ m/s$$

Step2: Analyze the vertical component of velocity at the landing point

The vertical motion of the ball is a free - fall motion with an initial vertical velocity \(v_{0y}=v_{0}\sin\theta\) and acceleration \(a = - g=- 9.8\ m/s^{2}\). Using the kinematic equation \(v_{y}^{2}-v_{0y}^{2}=2a\Delta y\). When the ball reaches the ground again, \(\Delta y = 0\). So \(v_{y}=-v_{0y}\) (the direction is opposite to the initial vertical direction).

$$v_{0y}=44.1\times\sin(34.4^{\circ})$$

$$v_{0y}=44.1\times0.565$$

$$v_{0y}\approx24.9\ m/s$$

So \(v_{y}=- 24.9\ m/s\)

Answer:

\(\vec{v}_{x}=36.4\frac{m}{s},\vec{v}_{y}=-24.9\frac{m}{s}\) (the first option)