Question

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.

Answer 1

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


`\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}`

Answer 2

After 20 seconds

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:

1. 
   `(d)/(dx)`x = 1
2. 
   `(d)/(dx)` 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