what is the max height

Question

A ball is thrown vertically upward. After t

 seconds, its height h

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

. After how long will it reach its maximum height?

Answer 1

Without using Calculus, you can also determine the maximum height and the time it takes to reach that height.

Remember that h(t) = −16t² + 80t is a quadratic meaning the graph forms a parabola. Since the leading coefficient is a negative, it is a parabola opening downward meaning it has a maximum value and that maximum value occurs at the vertex of the parabola.

To find the "x-value" of the vertex given by f(x) = ax² + bx + c, we use x = -b/(2a). In the case of h(t) = −16t² + 80t, a = -16 and b = 80 (c = 0 but that is of no value to us in this discussion). So to find the x-value of the vertex (in this case it is actually the t-value) we use t = -(80)/(2•(-16)) = -80/-32 = 2.5. So, the maximum height is reached after 2.5 seconds. To find what the height is, plug in t = 2.5:

h(t) = −16t² + 80t

h(2.5) = −16(2.5)² + 80(2.5) = 100 feet

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+200=100 ft= max height

-16t^2+80t

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

=-16(t-5/2)^2 +100 in vertex form

=a(x-h)^2+k

(h,k)=(5/2, 100)=vertex=maximum of

a downward opening parabola

(5/2 sec, 100 ft)