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

11-94.

Substitute 0 for *h*.

80*t* − 16*t*^{2} = 0

It is on the ground at 0 and 5 seconds.

Factor the equation.

16*t*(5 − *t*) = 0

When is the height positive?

D: 0 ≤ *t* ≤ 5

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

5 seconds

16*t* ^{2} − 80*t* + 64 = 0

Substitute 64 for *h* and solve for *t*.

Factor the equation.

(*t* − 1)(*t* − 4)

Simplify.

*t* ^{2} − 5*t* + 4

*t* = 1, 4

Solve for *t*.

Use these values to find height greater that 64 ft.

1< *t* < 4