Answer:
2.5 seconds , (50,2.5)
Step-by-step explanation:
Here we are given the equation of the path as
[tex]h=-8t^2+40t[/tex]
taking -8 out as GCF
[tex]h=-8(t^2-5t)[/tex]
adding and subtracting [tex]\frac{25}{4}[/tex] in the bracket
[tex]h=-8(t^2-5t+[tex]-\frac{25}{4})[/tex]
[tex]h=-8(t^2-5t+(\frac{5}{2})^2)+8\times\frac{25}{4}[/tex]
[tex]h=-8(t-\frac{5}{2})^2+50[/tex]
[tex](h-50)=-8(t-\frac{5}{2})^2[/tex]
Hence if we compare it with the standard equation of a parabola
we get that
vertex of the parabola formed above is
[tex](50,\frac{5}{2})[/tex]
Where h is on y axis and t is on x axis.
Hence the throw attains maximum height at t = 2.5 sec
and the coordinates of the maximum height attained will be (50,2.5)
