h(t)=−16t2+56t+32

Question

A person standing close to the edge on top of a 32-foot building throws a ball vertically upward. The quadratic function h(t)=−16t^2+56t+32 models the ball's height about the ground, h(t), in feet, tt seconds after it was thrown.

What is the maximum height of the ball?



How many seconds does it take until the ball hits the ground?

Answer 1

The maximum height occurs at the vertex of the parabola. That vertex occurs at t = -b/(2a) (for h(t) = at² + bt + c)

So a = -16 and b = 56

t = -56/(2(-16)) = 1.75 seconds

Plugging in t = 1.75, we get:

h(1.75) = -16(1.75)2 + 56(1.75) + 32 = 81

So the maximum height is 81 feet

The ball hits the ground at h = 0 so:

0 = −16t² + 56t + 32

0 = -8((2t² - 7t - 4)

0 = -8(2t + 1)(t - 4)

Setting each piece equal to zero, t = -1/2 and t = 4

We can ignore the t =-1/2 which would be 1/2 sec before time started) making the answer t = 4 seconds

Answer 2

h(t)=−16t² + 56t + 32

The equation is a parabola, the max height is at the vertex.

The vertex is on the line of symmetry, which is t = -b/2a

t=-(56)/(2*-16) = 1.75 secs

The Maximum height of the ball is h(1.75)

h(1.75)=-16(1.75)² + 56(1.75) +32 = 81 feet.

How many seconds does it take until the ball hits the ground?

Find t when h(t)=0

-16t²+56t+32=0

Solving the quadratic equations:

t=(-b±√(b²-4ac))/2a

t=(-56±√(56²-4(-16)(32)))/2(-16)

t=-0.5 and t=4

It takes 4 secs for the ball to hit the ground