Question

t (s) vx (m/s) vy (m/s) 0.0 16.0 0.0 1.0 16.0 -9.8 2.0 16.0 -19.6 3.0 16.0 -29.4 4.0 16.0 -39.2 \the objects acceleration is directed downward.\ \the objects horizontal velocity is constant; its vertical velocity is changing.\ arrows represent velocity vectors

Explanation:

Step1: Analyze horizontal velocity

From the table, $v_x$ is always $16.0$ m/s for different times $t$. So horizontal velocity is constant.

Step2: Analyze vertical velocity

The values of $v_y$ are changing with time ($0.0$, $- 9.8$, $-19.6$, $-29.4$, $-39.2$ m/s at $t = 0.0,1.0,2.0,3.0,4.0$s respectively). So vertical velocity is changing.

Step3: Analyze acceleration direction

Acceleration $a_y=\frac{\Delta v_y}{\Delta t}$. Since $v_y$ is decreasing in the positive - y direction (becoming more negative), the acceleration is in the negative - y direction (downward).

Answer:

The statements "The object's horizontal velocity is constant; it's vertical velocity is changing." and "The object's acceleration is directed downward." are both correct based on the data in the table.