A projectile is thrown upward so that its distance above the ground after t seconds is given by the function h(t) = -16t2 + 640t. After how many seconds does the projectile take to reach its maximum height? Show your work for full credit.

notice the picture below

the maximum height is reached at the vertex of the parabola

thus, just find the vertex, and the value for the y-axis(height in feet), is how high it went at that point, before it began falling back down

[tex]\bf \textit{vertex of a parabola}\\ \quad \\ y = {{ a}}x^2{{ +b}}x{{ +c}}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)\\\\ -----------------------------\\\\ \begin{array}{lccclll} h(t)=&-16t^2&+640t&+0\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}[/tex]

danieltirado danieltirado · Answer 2 · 2019-10-17T05:38:41+02:00

Answer:

After 20 seconds

Step-by-step explanation:

We have the function h(t) = -16t2 + 640t where t is the time in seconds. This function is a parable, and to get the maximum height of a parabola we have to find the first derivative of the function:

h'(t) = -32t + 640

To find the derivative we can do it by parts:

We have to use the derivative formulas:

[tex]\frac{d}{dx}[/tex]x = 1
[tex]\frac{d}{dx}[/tex] x^n = nx^n-1

So with this formulas we can get the derivative:

(d/dx)(-16t^2) = -32t

(d/dx)(640t) = 640

We know that the derivative that we just got (h'(t) = -32t + 640) is the same as the slope in a certain point, and we know that when a parable reaches the maximum height the slope is 0.

So we can make the expression equal to 0:

-32t + 640 = 0 and now we solve for t:

-32t = -640

t = -640/-32

t = 20 seconds