Question

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

Explanation:

Step1: Analyze the first graph (acceleration vectors)

In projectile motion, the acceleration is always \(a = g=- 9.8\space m/s^{2}\) (downward). The first graph shows acceleration vectors (red arrows) all pointing downward, which is correct for projectile - motion acceleration.

Step2: Analyze the second graph (velocity vectors)

In projectile motion, the horizontal component of velocity \(v_{x}\) is constant (\(a_{x}=0\)), and the vertical component of velocity \(v_{y}\) changes due to gravity (\(a_{y}=-g\)). The second graph has a non - vertical velocity vector at the top of the trajectory. At the maximum height of a projectile, the vertical component of velocity \(v_{y} = 0\), and the velocity vector should be horizontal. But in projectile motion, the acceleration is always vertical (\(a=-g\hat{y}\)). The non - vertical arrow in the velocity - vector graph (the arrow at the top of the trajectory in the right - hand graph) is incorrect in terms of representing projectile - motion velocity vectors.

Step3: Analyze the first table (displacement)

The horizontal displacement \(d_{x}=v_{0x}t\). If \(v_{0x} = 8\space m/s\), \(d_{x}\) values (\(0,8,16,24,32\)) for \(t = 0,1,2,3,4\space s\) are correct (\(d_{x}=v_{0x}t\)). The vertical displacement \(d_{y}=v_{0y}t-\frac{1}{2}gt^{2}\). If \(v_{0y} = 0\), \(d_{y}=-\frac{1}{2}(9.8)t^{2}\). For \(t = 1\space s\), \(d_{y}=- 4.9\space m\); for \(t = 2\space s\), \(d_{y}=-19.6\space m\); for \(t = 3\space s\), \(d_{y}=-44.1\space m\); for \(t = 4\space s\), \(d_{y}=-78.4\space m\), which is correct.

Step4: Analyze the second table (velocity)

The horizontal velocity \(v_{x}=v_{0x}\) (constant, \(v_{0x}=12\space m/s\)). The vertical velocity \(v_{y}=v_{0y}-gt\). If \(v_{0y} = 0\), \(v_{y}=-gt\). For \(t = 1\space s\), \(v_{y}=-9.8\space m/s\); for \(t = 2\space s\), \(v_{y}=-19.6\space m/s\); for \(t = 3\space s\), \(v_{y}=-29.4\space m/s\); for \(t = 4\space s\), \(v_{y}=-39.2\space m/s\), which is correct.

Answer:

The right - hand graph (the one with velocity vectors that has a non - vertical vector at the top of the trajectory)