Question

An object is thrown upward at a speed of 109 feet per second by a machine from a height of 7 feet off the ground. The height

h
h
of the object after
t
t
seconds can be found using the equation
h
=
−
16
t
2
+
109
t
+
7
h
=
-
16
t
2
+
109
t
+
7

Answer 1

Answer:

Explanation:
With the Given equation:
h = -16t^2 + 109t +7
When will the height be 167.
In other words at what time (t) will h be 167.
Plug in 167 for h in this equation.
This will result in:
167=-16t^2+109t+7
Rearrange
16t^2-109t+160 = 0
Use the Quadratic equation to solve for t.
t = (109 - sqrt(109^(2) - 4(-6)(160))/(2(16))
t = 4.67 seconds

When will the object reach the ground?
This time set h = 0.
0 = -16t^2 +109t +7
Use the Quadratic equation to solve for t.
t = (-109 - sqrt(109^(2) - 4(-16)(7))/(2(-16))
t = 6.876 seconds