Vertical Motion
Object Dropped
2
h  16t  s
s
Object Thrown Upward
2
h  16t  vt  s
h  height of object (feet)
t  time (seconds)
s  initial height (feet)
v
s
v  initial velocity (feet/second)
We use the quadratic formula to solve this
How long will a ball thrown upwards at 20 ft/sec
stay in the air if it is thrown from a 100 ft. cliff?
s  100ft.
v 20 ft./sec.
h  0 ft.
20
100
2
h  16t  vt  s
2
0  16t  20t  100
4
4
2
0  4t  5t  25
a  4
b  b  4ac b  5
2a
c  25
2
5  5  4-425
2-4
2
5  425 5  20.6

8
8
 1.95 or 3.2sec.
Objective - To graph quadratic inequalities.
2
Graph y  x  5
y
Test
x a ypoint inside
the parabola
– if it
2
-3 shade
3  5  4
works
2
inside,
if it
-2 2
2 doesn’t
 5  1
shade
-1 outside.
1  5  4
Test
0 (0,0)
0  5  5
2
2
 55  4
01  012 
2 2  5  1
2
25
0


3 3  5  4
True so shade inside
x
Vertex
(0,-5)
2
Graph y  x  x
y
x aypoint inside
Test
2 – if it
the
parabola
-3 3  3  12
2
works shade
-2 2  2  6
inside, if it doesn’t
2
-1 outside.
12 1  2
shade
0 0  0  0
Test (1,
2 0)
1 1 21  0
0 2 (21)2 2(2
1)
2
3 0
32  3  6
False so shade
outside
x