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 ($v_y$).

Explanation:

Step1: Horizontal speed

In projectile motion, the horizontal speed \(v_x\) remains constant (assuming no air resistance). So for points \(B\), \(D\), and \(E\), \(v_x\) is the same.

Step2: Vertical speed

The vertical motion of a projectile is affected by gravity (\(a = - g=- 9.8\ m/s^{2}\)). At the highest - point of the projectile's motion (\(v_y = 0\)).

• For point \(B\): The ball is moving upwards, so \(v_y>0\).
• For point \(D\): The ball is at a position above the highest - point (assuming the general projectile trajectory concept, if we consider the symmetry and the motion direction). The vertical velocity magnitude at a certain height on the ascending part is the same as the magnitude at the same height on the descending part. But since it's moving upwards at \(B\) and we assume the order of motion (ascending then descending), if we consider the motion from \(B\) to \(D\) (ascending further), but wait, no - actually, in a projectile, the vertical velocity at \(B\) (ascending) has a positive \(v_y\), at \(D\) (assuming it's on the ascending part before the peak, no - wait, no. Wait, in a projectile, the vertical velocity at \(B\) (lower position) is \(v_{yB}\), at \(D\) (higher position but not peak) \(v_{yD}=v_{yB}-gt\) (\(t = 1s\) interval). At \(E\) (descending part), \(v_{yE}\) has the same magnitude as \(v_{yB}\) but opposite in direction (\(v_{yE}<0\)). The magnitude of vertical velocity: \(|v_{yB}|>|v_{yD}|\) (because \(v_{yD}=v_{yB}-gt\) and \(v_{yB}>0\), \(t = 1s\), \(g = 9.8m/s^{2}\)), and \(|v_{yE}| = |v_{yB}|\). But if we consider the sign (where upward is positive), \(v_{yB}>v_{yD}>v_{yE}\) (since \(v_{yE}<0\), \(v_{yD}>0\) but less than \(v_{yB}\))

Answer:

For \(v_x\): All the Same. For \(v_y\): \(B > D> E\)