what is the max hieght

Question

A ball is thrown vertically upward. After t

 seconds, its height h

 (in feet) is given by the function =ht−80t16t2

. What is the maximum height that the ball will reach?

Do not round your answer.

Answer 1

The vertex of the parabola is found at a t value of -b/2a

For h(t) = -16t² + 80t; a = -16 and b = 80

-b/2a = -80/-32 = 2.5 sec.

Plugging 2.5 into the function for t will give you the max ht.

-16(2.5²) + 80(2.5) = -100 + 200 = 100 ft.

Answer 2

h(t) = -16t^2 + 80t

h'(t) = -32t +80 =0

t = 80/32 = 5/2 = 2 1/2 seconds to reach max height

h(5/2) = -16(5/2)^2 +80(5/2) = 100 feet = max height

but since you listed this under algebra 2 and not calculus

you may not yet understand the above calculation

so do an algebraic manipulation of the quadratic polynomial

rewrite it in vertex form. It's a downward opening parabola, if you graph it

-16t^2 +80t

=-16(t^2 -5t)

complete the square, add and subtract the square of half the coefficient of the linear term

= -16(t^2 -5t + (5/2)^2) + (16)(5/2)^2

= -16(t-5/2)^2 +16(25/4)

=- 16(t-2.5)^2 +100

that's in vertex form with the vertex = maximum point

a(x-h)^2 +k is the vertex form with (h,k) = the vertex

in this example, (2.5, 100) is the vertex, where 100 is the maximum height and 2.5 = the time when maximum height is reached

or you could just graph the polynomial and look at the maximum point of the graph

Desmos has an online graphing calculator, or use a handheld one

if you take a physics or engineering course, you'll see this basic quadratic equation

h(t) = a(x-h)^2 + k

or (a/2)t^2 + vot + ho

where vo= initial velocity at time t=0

ho = initial height

a = -32 feet per second per second = the effect of gravity at sea level

or a=-9.8 meters per second per second in the metric system