Question

one of these representations is not like the others. tap on the one that doesnt belong. \the vertical velocity changes by -9.8 m/s every second.\ \the objects horizontal velocity remains the same throughout its motion.\ check answer

Explanation:

Step1: Analyze the vertical velocity

The vertical velocity changes by -9.8 m/s every second (due to gravity), and the vertical velocity - time graph has a slope of -9.8 m/s/s (acceleration due to gravity \(g=- 9.8\ m/s^{2}\)). In projectile motion (assuming no air - resistance), the horizontal velocity remains constant.

Step2: Analyze the table

In the table, for the horizontal direction, \(d_{x}=v_{x}t\). When \(t = 1.0\ s\), \(d_{x}=12.0\ m\), so \(v_{x}=12\ m/s\); when \(t = 2.0\ s\), \(d_{x}=24.0\ m\) (\(v_{x}=\frac{24}{2}=12\ m/s\)); when \(t = 3.0\ s\), \(d_{x}=36.0\ m\) (\(v_{x}=\frac{36}{3}=12\ m/s\)); when \(t = 4.0\ s\), \(d_{x}=48.0\ m\) (\(v_{x}=\frac{48}{4}=12\ m/s\)). For the vertical direction, \(d_{y}=v_{y0}t+\frac{1}{2}at^{2}\). Assuming \(v_{y0} = 0\), \(a=-9.8\ m/s^{2}\), when \(t = 1.0\ s\), \(d_{y}=\frac{1}{2}\times(- 9.8)\times1^{2}=-4.9\ m\), but the table shows \(d_{y}=-9.8\ m\) at \(t = 1.0\ s\).

Answer:

The table (the one with \(t\), \(d_{x}\), \(d_{y}\))