## Intermediate Algebra (6th Edition)

The height of the object at $t=0$ is $576$ ft. The height of the object at $t=2$ is $672$ ft. The height of the object at $t=4$ is $640$ ft. The height of the object at $t=6$ is $480$ ft.
The given equation is $h(t)=-16t^{2}+80t+576$. To find the height of the object when $t=0$, substitute $0$ in for $t$. $h(0)=-16(0)^{2}+80(0)+576$ Evaluate: $-16(0)^{2}+80(0)+576=576$ Therefore, the height of the object when $t=0$ is $576$ ft. To find the height of the object when $t=2$, substitute $2$ in for $t$. $h(2)=-16(2)^{2}+80(2)+576$ Evaluate: $-16(2)^{2}+80(2)+576=672$. Therefore, the height of the object when $t=2$ is $672$ ft. To find the height of the object when $t=4$, substitute $4$ in for $t$. $h(4)=-16(4)^{2}+80(4)+576$ Evaluate: $-16(4)^{2}+80(4)+576=640$. Therefore, the height of the object when $t=4$ is $640$ ft. To find the height of the object when $t=6$, substitute $6$ in for $t$. $h(6)=-16(6)^{2}+80(6)+576$ Evaluate: $-16(6)^{2}+80(6)+576=480$. Therefore, the height of the object when $t=6$ is $480$ ft.