Question

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

Explanation:

Step1: Analyze the first table

In projectile motion, the horizontal displacement \(d_x = v_{0x}t\). From the first table, when \(t = 1.0\ s\), \(d_x=8.0\ m\), so \(v_{0x}=\frac{d_x}{t}=\frac{8.0\ m}{1.0\ s} = 8.0\ m/s\). The vertical displacement \(d_y=v_{0y}t-\frac{1}{2}gt^{2}\), with \(v_{0y} = 0\), \(d_y=-\frac{1}{2}(9.8\ m/s^{2})t^{2}\).

Step2: Analyze the second table

In projectile motion, the horizontal velocity \(v_x\) is constant (\(a_x = 0\)), and the vertical velocity \(v_y=v_{0y}-gt\). From the second table, \(v_x = 12.0\ m/s\) (constant), \(v_{0y} = 0\), and \(v_y=-gt\) (\(g = 9.8\ m/s^{2}\)).

Step3: Analyze the first graph

The first graph shows velocity vectors. In projectile motion, the horizontal velocity component should be constant. But according to the first table, \(v_{0x}=8.0\ m/s\), while from the second table \(v_{0x} = 12.0\ m/s\). The inconsistency is due to the first graph's horizontal velocity vector not matching the second - table's \(v_x\) value (the first table's \(v_{0x}\) calculated from \(d_x\) is \(8\ m/s\) which is different from the second table's \(v_x=12\ m/s\) which is the correct horizontal velocity in projectile motion as per the second - table's constant \(v_x\) and the acceleration - only - in - y - direction nature of projectile motion).

Answer:

The first graph (the one with velocity vectors)