Answer
The values found in part (b) are the same because $h(t)=-16(t^2-14)$ is just a factored form of $h(t)=-16t^2+224$, which means that they are equivalent to each other.
Work Step by Step
$h(t)=-16(t^2-14)$ is just a factored form of $h(t)=-16t^2+224$.
This means that −$-16(t^2-14)$ and $-16t^2+224$ are equivalent to each other.
Thus, when you evaluate $h(t)=-16t^2+224$ and $h(t)=-16(t^2-14)$ for t=2, the two expressions will give the same value.
