h(t)=−16t2+64t+80

Question

A person standing close to the edge on top of a 80-foot building throws a ball vertically upward. The quadratic function h(t)=−16t2+64t+80 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

Given:

t = time in seconds after the ball was thrown

h(t) = the height of the ball after t seconds.

h(t)=−16t²+64t+80

Question (1): What is the maximum height of the ball?

Remember that the graph of the function is a parabola open down because the leading coefficient is negative. Therefore, to get the maximum height we just have to find the vertex of this parabola. Since the function is in standard form h(t) = at²+bt+c, the formula in getting the vertex is:

V(h,k) = (-b/(2a), h(-b/(2a)))

-b/(2a) = -64/(2*(-16)) = 2 seconds (the time it reaches the highest point.)

h(-b/(2a)) = h(2) = -16(2)² + 64(2) + 80 = 144 feet (Maximum height)

Question (2): How many seconds does it take until the ball hits the ground?

If it hits the ground, it means height is zero. h(t)=0.

0= -16t² + 64t + 80

Factor:

0 = -16(t² - 4t - 5)

0 = -16(t - 5) (t + 1)

t-5 =0 or t+1=0

We eliminate t+1=0 because of negative value of t. Time should always be positive in this case.

t = 5 seconds to hit the gound.

Answer 2

h(t) = −16t² + 64t + 80

To find the maximum height of the ball, we need to find the vertex of the quadratic function.

x = -b/2a = -64/2(-16) = 2

h(2) = −16(2)² + 64(2) + 80 = -64 + 128 + 80 = 144 feet

We need to find how long it takes for the ball to reach the ground.

−16t² + 64t + 80 = 0

-16(t² - 4t - 5) = 0 (Factor by GCF of -16)

-16(t - 5)(t + 1) = 0 (Factor inside the parenthesis by box method, factoring groups, or diamond method)

t - 5 = 0 or t + 1 = 0 (Set each factor to zero by Zero Product Property)

t = 5 or t = -1 (Isolate the variable "t")

There is no such thing as a negative time. So, the ball will reach the ground at 5 seconds.