Question

An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.9t2 + 20t + 65. What is the object's maximum height?

Answer 1

You can find the maximum height of the object from the vertex. Do you remember how to find the vertex of an equation?

Answer 2

i really don't know i'm sorry i'm so bad at math

Answer 3

You can either complete the square or find the vertex by this formula
\(\Large Vertex=(h,k)=(\LARGE-\frac{b}{2a},c-\frac{b^2}{4a})\)
Where ax^2+bx+c=0 is the standard form
-4.9t^2 + 20t + 65=0
a=-4.9
b=20
c=65