Question

one of these representations is not like the others. tap on the one that doesnt belong.

the objects vertical velocity changes by the same amount every second.

Explanation:

Step1: Analyze the motion

In projectile motion, the horizontal velocity \(v_x\) is constant (no acceleration in \(x -\)direction, \(a_x = 0\)), and the vertical velocity \(v_y\) changes due to gravity (\(a_y=-g=- 9.8\ m/s^{2}\)). The displacement in \(x -\)direction is given by \(d_x=v_{0x}t\) (uniform motion) and in \(y -\)direction by \(d_y = v_{0y}t-\frac{1}{2}gt^{2}\).

Step2: Check the tables

• For the displacement table: \(d_x\) follows \(d_x = 8t\) (since when \(t = 1\ s\), \(d_x=8\ m\); \(t = 2\ s\), \(d_x = 16\ m\) etc., so \(v_{0x}=8\ m/s\)) and \(d_y\) follows \(d_y=-4.9t^{2}\) (since when \(t = 1\ s\), \(d_y=-4.9\ m\); \(t = 2\ s\), \(d_y=-19.6\ m=-4.9\times4\) etc., using \(d_y=v_{0y}t-\frac{1}{2}gt^{2}\), with \(v_{0y} = 0\)).
• For the velocity table: In projectile motion, \(v_x\) should be constant (\(v_x = 8\ m/s\) as per \(d_x\) relation) and \(v_y\) should change as \(v_y=v_{0y}-gt\). If \(v_{0y} = 0\), then \(v_y=-9.8t\). At \(t = 1\ s\), \(v_y=-9.8\ m/s\); \(t = 2\ s\), \(v_y=-19.6\ m/s\) etc. But the given velocity table has \(v_y=-9.8\ m/s\) for all \(t\), which is wrong.

Step3: Check the text

The text “The object’s vertical velocity changes by the same amount every second” is correct as \(a_y=-g\), \(\Delta v_y=a_y\Delta t\), and \(\Delta v_y\) is constant (\(\Delta v_y=-9.8\ m/s\) per second). The diagram shows acceleration vectors (only vertical, which is correct for projectile motion as \(a_x = 0\), \(a_y=-g\)).

Answer:

The velocity table (the table with \(v_x\) and \(v_y\) columns)