(ALGEBRA 1)
Answer:
43.06 ft
3.20 seconds
Step-by-step explanation:
h = -16t^2 +50t +4
The maximum is at the vertex
The x coordinate for the vertex is a the axis of symmetry
The axis of symmetry is
h=-b/2a when the equation is at^2 +bt+c
h = -50/2(-16)
=-50/-32
= 1.5625
The maximum is when x is at 1.5625
to find y we put this into the equation
h = -16(1.5623)^2 + 50(1.5625) +4
= -16(2.44140625) + 78.125+4
= -39.0625+78.125+4
=43.0625
To the nearest hundredth
43.06 ft
The second part is to find when it hits the ground, or when h=0
0 = -16t^2 +50t +4
Using the quadratic formula
-b±sqrt(b^2 -4ac)
------------------------
2a
-50±sqrt(50^2 -4*-16*4)
------------------------
2*-16
-50±sqrt(2500 +256)
------------------------
-32
-50±sqrt(2756)
------------------------
-32
-50 + sqrt(2756) or -50 + sqrt(2756)
----------------------- -----------------------
-32 -32
-.078 3.20
Since time cannot be negative
3.20 seconds
(b) y/3 + y/4 + y/5 = y/3 + y/4 + y/5 + 1
(c) z/3 + z/4 + z/5 = z/4 + z/5 + z/6 + 1$