A ball is thrown vertically upward. After t seconds, it's height h (in feet) is given by the function h(t)=40t-16t^2.

Question

math 105 (college course)

second part: What is the maxium height that the ball will reach?

Answer 1

What was the first part??

 

h(t) = -16t² + 40t

 

This is a parabola opening downward and the maximum

height will be at the vertex

 

The t coordinate of the vertex is -b/2a from quadratic equation

 

-b/2a = -40/-32 = 5/4 = 1.25

 

h(1.25) = _______

 

Plug this value in for t and solve the equation to find maximum height.

Answer 2

C-B^2/(4A)

 

Finds the max or min value. Just a little quicker

C=0 -40^2/(4(-16)) 1600/64

Answer 3

Use the formula

 

k = -b / (2a)   ----->  x-coordinate of the maximum.

 

where:
a = -16

b = 40

 

Maximum height is h(k).