Question

one of these representations is not like the others. tap on the one that doesnt belong.
arrows
represent
velocity
vectors
arrows
represent
acceleration
vectors

Explanation:

Step1: Analyze the velocity - vector diagram

In projectile motion, the horizontal component of velocity \(v_x\) is constant (assuming no air - resistance), and the vertical component of velocity \(v_y\) changes due to gravity. In the velocity - vector diagram (left - hand side upper diagram), the horizontal velocity vectors should be of the same length (since \(v_x=\text{constant}\)), but in the given velocity - vector diagram, the lengths of the horizontal velocity vectors are not the same.

Step2: Analyze the acceleration - vector diagram

In projectile motion, the acceleration is \(a = g=- 9.8\space m/s^{2}\) (downward). In the acceleration - vector diagram (right - hand side upper diagram), all the acceleration vectors are of the same length (since \(a = g=\text{constant}\) in the vertical direction and \(a_x = 0\)).

Step3: Analyze the position - time table

From the position - time table (\(d_x=v_{0x}t\), \(d_y = v_{0y}t-\frac{1}{2}gt^{2}\)). If \(v_{0x}=8\space m/s\) (from \(d_x\) values: when \(t = 1s\), \(d_x=8m\)), and \(v_{0y} = 0\space m/s\) (\(d_y=-\frac{1}{2}gt^{2}\), when \(t = 1s\), \(d_y=-4.9m\), \(g = 9.8m/s^{2}\)).

Step4: Analyze the velocity - time table

From the velocity - time table, \(v_x=\text{constant}=12m/s\) (which is a contradiction with the position - time table where \(v_{0x}\) seems to be \(8m/s\), but more importantly, looking at the nature of vectors and tables). The key is the velocity - vector diagram. In projectile motion, the horizontal velocity component is constant. The velocity - vector diagram (left - hand side) has non - constant horizontal velocity vectors.

Answer:

The velocity - vector diagram (the left - hand side upper diagram) doesn't belong.