Height of a ball

Question

If a ball is thrown directly upward with a velocity of 40ft/s, it's height (in feet) after t seconds is given by y=40t-16t^2. What is the maximum height attained by the ball? 

Answer 1

The keyword in the question is "maximum", this tells you that you are looking for the vertex,(turning point of the graph, the highest point of the of the parabola). Since they are asking for the maximum height, and "t" represents seconds, you must find the "y" value.

Standard quadratic form is:
ax^2+bx+c
Use the vertex formula:
x=-b/2a
to find the "x" value of the vertex, then plug that value into the original equation as a substitute for "x".

Given y= 40t-16t^2
a=16, b=40
x=-40/2(-16)
x= 5/4 (or 1.25)
So,
y= 40(1.25) - 16(1.25)^2
y = 25

The maximum height attained is 25 ft