If a baseball is projected upward from ground level with an initial velocity of 32 feet per second, then its height is a function of time, given by s= -16t^2 + 32t. What is the maximum height reached by the ball?

Question

anonymous · Answer

so this is just a quadratic equation.

anonymous · Answer

you know from your formulas (or from calculus), that the vertex of a quadratic equation is at    -b/2a

anonymous · Answer

where the quadratic is in the form:
$$ax^2 +bx +c$$

anonymous · Answer

so b is 32 and a is -16 (they are the coefficients) (by the way c is 0)

anonymous · Answer

you can now use the formula to find the vertex

anonymous · Answer

well, its x coordinate at least

anonymous · Answer

ohh! I understand now! thank u @MrMoose

anonymous · Answer

you are welcome

lgbasallote · Answer

if you're interested in using derivatives...the derivative of distance is the velocity

s' = -32t + 32

it is maximum height when s' = 0 so

0 = -32t + 32

32t = 32

solve for t