Step-by-step explanation:
A ball is thrown starting at a time of 0 and a height of 2 meters. The height of the ball follows the function as follows:
[tex]H(t)=-4.9t^2+25t+2[/tex]
We need to find the height of the ball from 0 to 5 seconds
At t = 0 s
[tex]H(0)=-4.9t^2+25t+2\\\\=-4.9(0)^2+25(0)+2\\\\=2\ m[/tex]
At t = 1 s
[tex]H(1)=-4.9t^2+25t+2\\\\=-4.9(1)^2+25(1)+2\\\\=22.1\ m[/tex]
At t = 2 s
[tex]H(2)=-4.9t^2+25t+2\\\\=-4.9(2)^2+25(2)+2\\\\=32.4[/tex]
At t = 3 s
[tex]H(3)=-4.9t^2+25t+2\\\\=-4.9(3)^2+25(3)+2\\\\=32.9\ m[/tex]
At t = 4 s
[tex]H(4)=-4.9t^2+25t+2\\\\=-4.9(4)^2+25(4)+2\\\\=23.6\ m[/tex]
At t = 5 s
[tex]H(5)=-4.9t^2+25t+2\\\\=-4.9(5)^2+25(5)+2\\\\=4.5\ m[/tex]
t = 0 1 2 3 4 5
H = 2 22.1 32.4 32.9 23.6 4.5
