Question

a ball is launched at an angle. its location at 1 - s intervals is shown. rank the three indicated locations based on their horizontal speed (v_x) and vertical speed (v_y). tap table cells to toggle through answer options. tap here if all the same tap here if all the same

Explanation:

Step1: Recall projectile - motion properties

In projectile motion, the horizontal speed ($v_x$) is constant in the absence of air - resistance. So, $v_{xB}=v_{xD}=v_{xE}$.

Step2: Analyze vertical speed

The vertical speed $v_y$ changes due to gravity. At the highest point of the projectile's path, $v_y = 0$. Point E is at the highest point among B, D, and E, so $v_{yE}=0$. Point B is on the way up, so $v_{yB}>0$. Point D is on the way down, so $v_{yD}<0$. And $|v_{yD}| > v_{yB}$ (because it has fallen further). So, $v_{yB}>v_{yE}>v_{yD}$.

Answer:

For $v_x$: Tap here if All the Same
For $v_y$: $v_{yB}>v_{yE}>v_{yD}$