Question

21 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?

\\( \vec { v _ { x } } = - 5.27 \mathrm { m } / \mathrm { s } , \vec { v _ { y } } = 14.8 \mathrm { m } / \mathrm { s } \\)
\\( \vec { v _ { x } } = 14.8 \mathrm { m } / \mathrm { s } , \vec { v _ { y } } = - 5.27 \mathrm { m } / \mathrm { s } \\)
\\( \vec { v _ { x } } = 5.27 \mathrm { m } / \mathrm { s } , \vec { v _ { y } } = - 14.8 \mathrm { m } / \mathrm { s } \\)
\\( \vec { v _ { x } } = 14.8 \mathrm { m } / \mathrm { s } , \vec { v _ { y } } = 5.27 \mathrm { m } / \mathrm { s } \\)

Explanation:

Step1: Calculate the x - component of velocity

The x - component of velocity \(v_x\) for a projectile is given by \(v_x = v\cos\theta\), where \(v = 15.8\ m/s\) and \(\theta=19.5^{\circ}\).

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

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

Since there is no acceleration in the x - direction (assuming no air resistance), the x - component of velocity remains constant throughout the motion.

Step2: Calculate the y - component of velocity

The y - component of velocity \(v_y\) for a projectile is given by \(v_y=-v\sin\theta\) (negative because the balloon is moving downwards just before hitting the ground).

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

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

Answer:

\(\overrightarrow{v_x}=14.8\ m/s,\overrightarrow{v_y}=-5.27\ m/s\) (the second option)