Question

one of these representations is not like the others. tap on the one that doesnt belong.
\the objects vertical acceleration changes by - 9.8 m/s every second.\
arrows represent velocity vectors
arrows represent force vectors
check answer

Explanation:

Step1: Analyze the nature of projectile motion

In projectile motion, the vertical acceleration \(a_y=-g = - 9.8\space m/s^{2}\) (constant). The change in vertical velocity \(\Delta v_y=a_y\Delta t\). The horizontal velocity \(v_x\) is constant (assuming no air - resistance).

Step2: Analyze each representation

• Table: The table shows \(v_x = 8.0\space m/s\) (constant) and \(v_y\) changes by \(\Delta v_y=-9.8\space m/s\) each second (\(\Delta t = 1s\)). Since \(a_y=\frac{\Delta v_y}{\Delta t}=-9.8\space m/s^{2}\) (constant).
• Velocity - vector diagram: The horizontal component of velocity is constant (arrows of the same length in the x - direction), and the vertical component of velocity changes by a constant amount (\(-9.8\space m/s\) per second, as the vertical arrows change their length by the same amount each step).
• Force - vector diagram: In projectile motion (neglecting air resistance), the only force is \(F = mg\) (weight), so the force is constant (\(F_y=-mg\), \(F_x = 0\)). The acceleration \(a=\frac{F}{m}\), so \(a_y=-g=-9.8\space m/s^{2}\) (constant), \(a_x = 0\).
• Text statement: The text says “The object's vertical acceleration changes by \(-9.8\space m/s\) every second”. This is wrong. Acceleration is \(-9.8\space m/s^{2}\) (constant), and the change in velocity \(\Delta v=a\Delta t\). The unit of acceleration is \(m/s^{2}\), and the unit of the change in velocity is \(m/s\).

Answer:

The text statement “The object's vertical acceleration changes by \(-9.8\space m/s\) every second” doesn't belong.