You throw a nerf basketball of mass 0.002 kg into the air. At $t = 0$, the ball leaves your hand. At $t = 0.10$ s, the net force on the ball is $(-0.0177, -0.0217, 0)$ N, its velocity is $(1.87, 0.84, 0)$ m/s, and its position is $(0.325, 0.110, 0)$ m. (Besides the gravitational force by Earth, air also exerts a force on the ball.)
Part 1
Your answer is partially correct.
What will be its position at $t = 0.12$ s?
$\vec{r} = \langle$ 0.32146 , 0.10566 , 0 $\rangle$ m