Question

A projectile is thrown upward so that its distance above the ground after t seconds is given by the function h(t) = -16t² + 704t. After how many seconds does the projectile take to reach its maximum height? Show your work for full credit.

Answer 1

22 secs.

Explanation:

h(t) = -16t^2 + 704t

The maximum will happen where the first derivative is equal to 0.

h’(t) = -32t + 704

-32t + 704 = 0 => -32t = -704 After subtracting 704 from both sides of the equation

Now divide both sides of the equation by -32

t = 22 secs.

Double Check: recalculated/reasonable ✅ ✅

Answer: 22 secs.

Just an interesting aside. After 22 secs, the projectile will be -16(22)^2 + 704(22) = -7744 + 15,488 = 7744 units up as its maximum height.

Answer 2

h(t) = -16t^2 +704t = -16(t^2 -44t)
Putting in vertex form: -16(t^2 -44t +484) + 704
h(t) = -16(t-22)^2 + 704
It takes 22 seconds to reach the maximum height, which is the vertex of the parabola at (22,704)