h(t)=−16t2+96t+32

Question

A baseball player throws a ball into the air with an initial vertical velocity of 96 feet per second. The ball is released at a height of 32 feet. The equation h(t)=−16t2+96t+32 models the height (in feet) of the ball as a function of the time (in seconds) after it is thrown.

What is the maximum height the ball reaches in feet?

At what time does the ball reach its maximum height in seconds?

Answer 1

Well based on the topics you put it under, I will be solving it using algebra.

h(t)= −16t^2 + 96t + 32 is a quadratic equation. This is f(x) = ax^2 + bx + c.

It will form a parabola. The x-axis is time and the y-axis is height.

The maximum height is the vertex of the parabola. The way to the coordinate, we will need to find the x-coordinate first (which is in the opposite order of your questions but it is the way).

x-coordinate of vertex = (-b)/(2*a)

Now from the equation, we can see that a = -16 and b = 96. So substituting those values in,

(-96)/ (2 * (-16)) = 96/32 = 3.

So it takes 3 seconds to reach maximum height.

Now to find the height, we simply substitute t = 3 into h(t).

h(3) = −16 (3) ^2 + 96(3) + 32

= -144 + 288 + 32

= 176

So the maximum height is 176 feet.

Answer 2

h = -16t^2 + 96t +32

h' = -32t + 96 = 0

t = 3 seconds to reach max height

h(3) = -16(3)^2 + 96(3) + 32 = 176 feet = max height