
Home > A2C > Chapter 11 > Lesson 11.2.4 > Problem 11-94
11-94.
Substitute 0 for h.
80t − 16t2 = 0
It is on the ground at 0 and 5 seconds.
Factor the equation.
16t(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
16t 2 − 80t + 64 = 0
Substitute 64 for h and solve for t.
Factor the equation.
(t − 1)(t − 4)
Simplify.
t 2 − 5t + 4
t = 1, 4
Solve for t.
Use these values to find height greater that 64 ft.
1< t < 4