May I have help solving this problem?

Question

Suppose that an object is projected in such a way that its height above the ground in feet after t seconds i s given s(t)= 16t^2 +48t + 40. Find the maxium height of the object.

Answer 1

The equation must be:

 

s(t) = -16t² + 48t + 40

 

Otherwise it would have no maximum height value. 

 

This modified equation is the equation of a parabola.  Since the coefficient of the t² term is negative (-16), the parabola opens downward and its vertex will be the highest point on the parabola.  For any parabola of the form at² + bt + c, the t-coordinate of the vertex is -b/2a.  In your problem, b= 48, a = -16

 

t = -b/2a = (-48)/(2)(-16)

 

Solve for t, then plug that value of t into the s(t) equation to find the height (s) of the vertex.