2. Vertical Motion
• Compares the height of an object with the
time in flight.
2
1
0
0
2
h(t ) = − gt + v t + h
g = force of gravity:
32ft/sec or 9.8 m/sec
vo = initial velocity
ho = initial height
Jeff Bivin -- LZHS
3. A ball is thrown into the air from the top
of a 200 foot tall building with an initial
upward velocity of 80 ft/sec.
•
•
•
•
•
•
•
What is the maximum height of the ball?
When will the ball reach the maximum height?
When will the ball return to the ground?
When will the ball be at a height of 250 feet?
When will the ball be at a height of 400 feet?
When will the ball be at a height of 50 feet?
If the ball lands in a 20 foot deep pit, when will the ball hit
the bottom of the pit?
• What will be the height of the ball in 3 seconds?
• How far from the building will the ball land?
Jeff Bivin -- LZHS
4. A ball is thrown into the air from the top
of a 200 foot tall building with an initial
upward velocity of 80 ft/sec.
• What is the maximum height of the ball?
We need to use:
h(t ) = − 1 gt 2 + v0t + h0
2
g = 32 ft/s2
vo = 80 ft/s
ho = 200 ft
h(t ) = − (32)t + 80t + 200
h(t ) = − 16t 2 + 80t + 200
1
2
Jeff Bivin -- LZHS
2
5. A ball is thrown into the air from the top
of a 200 foot tall building with an initial
upward velocity of 80 ft/sec.
• What is the maximum height of the ball?
h(t ) = − 16t 2 + 80t + 200
Where is the maximum?
t
t
t
t
=
=
=
=
−b
2a
−80
2 ( −16 )
−80
−32
5
2
Jeff Bivin -- LZHS
h(
Find the vertex……
) = − 16( ) + 80( 5 ) + 200
2
h( 5 ) = − 16( 25 ) + 80( 5 ) + 200
2
4
2
h( 5 ) = 300
2
( t , h(t ) )
Vertex is: ( 5 , 300 )
2
5
2
5 2
2
300 ft.
6. A ball is thrown into the air from the top
of a 200 foot tall building with an initial
upward velocity of 80 ft/sec.
• When will the ball reach the maximum height?
2.5 sec
h(t ) = − 16t 2 + 80t + 200
Where is the maximum?
t
t
t
t
=
=
=
=
−b
2a
−80
2 ( −16 )
−80
−32
5
2
Jeff Bivin -- LZHS
h(
Find the vertex……
) = − 16( ) + 80( 5 ) + 200
2
5
5 2
h( 2 ) = − 16( 2 ) + 80( 5 ) + 200
2
h( 5 ) = 300
2
( t , h(t ) )
Vertex is: ( 5 , 300 )
2
5
2
5 2
2
7. A ball is thrown into the air from the top
of a 200 foot tall building with an initial
upward velocity of 80 ft/sec.
• When will the ball return to the ground? 6.830 sec.
h(t ) = − 16t 2 + 80t + 200
What is the height at the ground?
h(t) = 0
0 = − 16t + 80t + 200
2
t=
t=
−80 ± 80 2 − 4 ( −16 )( 200 )
2 ( −16 )
−80 ± 19200
−32
Get the decimal approximations:
Jeff Bivin -- LZHS
t ≈ − 1.830
t ≈ 6.830
8. A ball is thrown into the air from the top
of a 200 foot tall building with an initial
upward velocity of 80 ft/sec.
• When will the ball be at a height of 250 feet?
2
h(t ) = − 16t + 80t + 200 0.732 sec
250 feet
What height?
h(t) = 250
&
2
4.268 sec.
250 = − 16t + 80t + 200
0 = − 16t + 80t − 50
2
t=
t=
−80 ± 80 2 − 4 ( −16 )( −50 )
2 ( −16 )
−80 ± 3200
−32
Get the decimal approximations:
Jeff Bivin -- LZHS
t ≈ 0.732
t ≈ 4.268
9. A ball is thrown into the air from the top
of a 200 foot tall building with an initial
upward velocity of 80 ft/sec.
• When will the ball be at a height of 400 feet?
never
h(t ) = − 16t 2 + 80t + 200
What height?
h(t) = 400
400 = − 16t 2 + 80t + 200
0 = − 16t 2 + 80t − 200
t=
t=
−80 ± 80 2 − 4 ( −16 )( − 200 )
2 ( −16 )
−80 ± −6400
−32
Wait, what was the maximum height?
Jeff Bivin -- LZHS
300 ft
10. A ball is thrown into the air from the top
of a 200 foot tall building with an initial
upward velocity of 80 ft/sec.
• When will the ball be at a height of 50 feet?
6.453 sec.
h(t ) = − 16t 2 + 80t + 200
What height?
h(t) = 50
50 = − 16t 2 + 80t + 200
0 = − 16t 2 + 80t + 150
t=
t=
−80 ± 80 2 − 4 ( −16 )( 150 )
2 ( −16 )
−80 ± 16000
−32
Get the decimal approximations:
Jeff Bivin -- LZHS
t ≈ − 1.453
t ≈ 6.453
11. A ball is thrown into the air from the top
of a 200 foot tall building with an initial
upward velocity of 80 ft/sec.
• If the ball lands in a 20 foot deep pit, when will
h(t ) the ball 2hit thet bottom of the pit?
= − 16t + 80 + 200
What height?
h(t) = -20
− 20 = − 16t 2 + 80t + 200
20 feet below the ground
Jeff Bivin -- LZHS
12. A ball is thrown into the air from the top
of a 200 foot tall building with an initial
upward velocity of 80 ft/sec.
• If the ball lands in a 20 foot deep pit, when will
the ball hit the bottom of the pit?
6.972 sec.
2
h(t ) = − 16t + 80t + 200
What height?
h(t) = -20
− 20 = − 16t 2 + 80t + 200
0 = − 16t 2 + 80t + 220
t=
t=
−80 ± 80 2 − 4 ( −16 )( 220 )
2 ( −16 )
−80 ± 20480
−32
Get the decimal approximations:
Jeff Bivin -- LZHS
t ≈ − 1.972
t ≈ 6.972
13. A ball is thrown into the air from the top
of a 200 foot tall building with an initial
upward velocity of 80 ft/sec.
• What will be the height of the ball in 3 seconds?
296 ft.
h(t ) = − 16t 2 + 80t + 200
What time?
t=3
h(3) = − 16(3) 2 + 80(3) + 200
h(3) = − 144 + 240 + 200
h(3) = 296
Jeff Bivin -- LZHS
14. A ball is thrown into the air from the top
of a 200 foot tall building with an initial
upward velocity of 80 ft/sec.
• How far from the building will the ball land?
h(t ) = − 16t + 80t + 200
2
Wait !!!!
This formula compares time with
height, not horizontal distance.
Answer:
Jeff Bivin -- LZHS
we don’t know!
16. A diver dives off a 3 meter diving board
into a pool with an initial upward
velocity of 3.5 m/sec.
• What is the maximum height of the diver?
• When will the diver reach his/her maximum height?
• When will the diver splash into the water?
• What will be the height of the diver in 1 second?
Jeff Bivin -- LZHS
17. A diver dives off a 3 meter diving board
into a pool with an initial upward
velocity of 3.5 m/sec.
• What is the maximum height of the diver?
3.625 meters
• When will the diver reach his/her maximum height?
0.358 sec.
• When will the diver splash into the water?
1.217 sec.
• What will be the height of the diver in 1 second?
1.6 meters
Jeff Bivin -- LZHS
19. A taxi service operates between two airports transporting
200 passengers a day. The charge is $15.00. The owner
estimates that 10 passengers will be lost for each $2
increase in the fare. What charge would be most profitable
for the service? What is the maximum income? VERTEX
Define the variable
x = number of $2
price increases
f(x) = income
x=
−b
2a
x=
−250
2 ( −20 )
x = 6.25
Income = Price ● Quantity
f(x) = ( 15 + 2x ) ( 200 – 10x )
f(x) = 3000 – 150x + 400x – 20x2
f(x) = – 20x2 + 250x + 3000
f(6.25) = – 20(6.25)2 + 250(6.25) + 3000
Vertex is:
f(6.25) = 3781.25
( 6.25,
3781.25)
So, price = (15 + 2x) = (15 + 2(6.25)) = 15 + 12.5 = $27.50
Maximum income = f(x) = $3781.25
Jeff Bivin -- LZHS
