Net displacement via integration of velocity

An antiderivative of $v(t) = 20 - 9.8t$ is $s(t) = 20t - 4.9t^2$, since differentiating gives back $v(t)$. By the Net Change Theorem, the net displacement equals $s(4) - s(0)$.

Compute $s(4) = 20(4) - 4.9(4)^2 = 80 - 78.4 = 1.6$, and $s(0) = 0$. So the net displacement is $1.6 - 0 = 1.6$ meters. Since this value is positive, the ball ends up $1.6$ meters higher than where it started, even though it went up and came back down partway.

The ball is above its starting height at time four. This follows because the net displacement computed in part a, the integral of velocity from zero to four, equals positive one point six meters. A positive net displacement means the final position is higher than the starting position, even though the ball rose and partly fell back down in between.