Question

1. patton launches a water balloon at a speed of 15.8 meters per second in the positive x - direction at an angle of 19.5° above horizontal. what are the components of the velocity vector right before the balloon strikes the ground?

\begin{align}&overrightarrow{v_{x}}=-5.27 m/s,overrightarrow{v_{y}} = 14.8 m/s\\&overrightarrow{v_{x}}=14.8 m/s,overrightarrow{v_{y}}=-5.27 m/s\\&overrightarrow{v_{x}}=5.27 m/s,overrightarrow{v_{y}}=-14.8 m/s\\&overrightarrow{v_{x}}=14.8 m/s,overrightarrow{v_{y}}=5.27 m/send{align}

Explanation:

Step1: Analyze the horizontal component

In projectile motion, the horizontal component of velocity \(v_x\) remains constant (neglecting air - resistance). The formula for the horizontal component of velocity is \(v_x = v\cos\theta\), where \(v = 15.8\ m/s\) and \(\theta=19.5^{\circ}\).

$$v_x=15.8\cos(19.5^{\circ})$$

$$v_x = 15.8\times0.943 = 14.8\ m/s$$

Step2: Analyze the vertical component

The initial vertical component of velocity is \(v_{0y}=v\sin\theta\), and the final vertical component of velocity \(v_y\) before hitting the ground. Using the kinematic equation \(v_y^{2}=v_{0y}^{2}- 2g\Delta y\). Since the balloon starts and ends at the same height \(\Delta y = 0\) (assuming the launch and landing heights are the same), and using the conservation of mechanical energy (or kinematic equations with \(a=-g\)), \(v_y=-v\sin\theta\).

$$v_y=-15.8\sin(19.5^{\circ})$$

$$v_y=-15.8\times0.334=- 5.27\ m/s$$

Answer:

\(\vec{v}_x = 14.8\ m/s,\vec{v}_y=-5.27\ m/s\) (the second option)