### Home > A2C > Chapter 11 > Lesson 11.2.4 > Problem11-94

11-94.

For an object shot into the air, its height $h$ in feet above the ground after t seconds is given by the equation $h = 80t − 16t^{2}$. Use this equation to answer the following questions.

1. For what times is the object on the ground?

Substitute 0 for h.

80t − 16t2 = 0

It is on the ground at 0 and 5 seconds.

Factor the equation.

16t(5 − t) = 0

2. For what domain is this function reasonable?

• When is the height positive?

D: 0 ≤ t ≤ 5

1. How long did it take the object to hit the ground?

• At what time will the object be on the ground, not including when it starts?

• 5 seconds

1. For what times is the height greater than $64$ feet?

• $16t ^{2} − 80t + 64 = 0$

• Substitute 64 for h and solve for t.

• Factor the equation.

• $\left(t − 1\right)\left(t − 4\right)$

• Simplify.

• $t ^{2} − 5t + 4$

• $t = 1, 4$

• Solve for t.

• Use these values to find height greater that 64 ft.

• $1\lt t \lt 4$