Given:
[tex]y=-16t^2+vt+h[/tex]
Let's find the height of the ball, y, when:
t = 2 seconds
h = 20 ft
v = 72 ft/s
Plug in the values in the equation and solve for y.
We have:
[tex]\begin{gathered} y=-16(2)^2+72(2)+20 \\ \\ y=-16(4)+144+20 \\ \\ y=-64+144+20 \\ \\ y=100 \end{gathered}[/tex]
Therefore, the height of the ball is 100 ft.
ANSWER:
100 ft.
