Horizontal motion formula:
d = rt
distance = rate • time
Horizontal motion formula:
d = rt
distance = rate • time
Vertical motion formula:
d = rt – 5t2
Horizontal motion formula:
d = rt
distance = rate • time
Vertical motion formula:
d = rt – 5t2
d = distance in meters above or below
starting point
Horizontal motion formula:
d = rt
distance = rate • time
Vertical motion formula:
d = rt – 5t2
d = distance in meters above or below
starting point
r = initial upward velocity in meters per second
Horizontal motion formula:
d = rt
distance = rate • time
Vertical motion formula:
d = rt – 5t2
d = distance in meters above or below
starting point
r = initial upward velocity in meters per second
t = time in seconds since object was released
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
d = rt – 5t2
A person standing on the ground throws a ball
upward with an initial velocity of 40 m/sec.
sec.
1
2
3
4
5
6
7
8
calculation
distance
d = rt – 5t2
A person standing on the ground throws a ball
upward with an initial velocity of 40 m/sec.
sec.
calculation
1
2
3
4
5
6
7
8
40•1–5•1=40–5
distance
35 m
d = rt – 5t2
A person standing on the ground throws a ball
upward with an initial velocity of 40 m/sec.
sec.
calculation
1
2
3
4
5
6
7
8
40•1–5•1=40–5
40•2–5•4=80–20
distance
35 m
60 m
d = rt – 5t2
A person standing on the ground throws a ball
upward with an initial velocity of 40 m/sec.
sec.
calculation
1
2
3
4
5
6
7
8
40•1–5•1=40–5
40•2–5•4=80–20
40•3–5•9=120–45
distance
35 m
60 m
75 m
d = rt – 5t2
A person standing on the ground throws a ball
upward with an initial velocity of 40 m/sec.
sec.
calculation
1
2
3
4
5
6
7
8
40•1–5•1=40–5
40•2–5•4=80–20
40•3–5•9=120–45
40•4–5•16=160–80
distance
35 m
60 m
75 m
80 m
d = rt – 5t2
A person standing on the ground throws a ball
upward with an initial velocity of 40 m/sec.
sec.
calculation
1
2
3
4
5
6
7
8
40•1–5•1=40–5
40•2–5•4=80–20
40•3–5•9=120–45
40•4–5•16=160–80
40•5–5•25=200–125
distance
35 m
60 m
75 m
80 m
75 m
d = rt – 5t2
A person standing on the ground throws a ball
upward with an initial velocity of 40 m/sec.
sec.
calculation
1
2
3
4
5
6
7
8
40•1–5•1=40–5
40•2–5•4=80–20
40•3–5•9=120–45
40•4–5•16=160–80
40•5–5•25=200–125
40•6–5•36=240–180
distance
35 m
60 m
75 m
80 m
75 m
60 m
d = rt – 5t2
A person standing on the ground throws a ball
upward with an initial velocity of 40 m/sec.
sec.
calculation
1
2
3
4
5
6
7
8
40•1–5•1=40–5
40•2–5•4=80–20
40•3–5•9=120–45
40•4–5•16=160–80
40•5–5•25=200–125
40•6–5•36=240–180
40•7–5•49=280–245
distance
35 m
60 m
75 m
80 m
75 m
60 m
35 m
d = rt – 5t2
A person standing on the ground throws a ball
upward with an initial velocity of 40 m/sec.
sec.
calculation
1
2
3
4
5
6
7
8
40•1–5•1=40–5
40•2–5•4=80–20
40•3–5•9=120–45
40•4–5•16=160–80
40•5–5•25=200–125
40•6–5•36=240–180
40•7–5•49=280–245
40•8–5•64=320–320
distance
35 m
60 m
75 m
80 m
75 m
60 m
35 m
0m
35 m
1 sec
60 m
2 sec
35 m
1 sec
75 m
3 sec
60 m
2 sec
35 m
1 sec
80 m in 4 seconds
75 m
3 sec
60 m
2 sec
35 m
1 sec
80 m in 4 seconds
75 m
3 sec
60 m
2 sec
35 m
1 sec
75 m
5 sec
80 m in 4 seconds
75 m
3 sec
75 m
5 sec
60 m
2 sec
60 m
6 sec
35 m
1 sec
80 m in 4 seconds
75 m
3 sec
75 m
5 sec
60 m
2 sec
60 m
6 sec
35 m
1 sec
35 m
7 sec
80 m in 4 seconds
75 m
3 sec
75 m
5 sec
60 m
2 sec
60 m
6 sec
35 m
1 sec
35 m
7 sec
on the ground in 8 sec
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit
the ground?
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit
the ground?
d = rt – 5t2
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit
the ground?
d = rt – 5t2
0 = 40t – 5t2
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit
the ground?
d = rt – 5t2
0 = 40t – 5t2
0 = t(40 – 5t)
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit
the ground?
d = rt – 5t2
0 = 40t – 5t2
0 = t(40 – 5t)
t = 0 or 40 – 5t = 0
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit
the ground?
d = rt – 5t2
0 = 40t – 5t2
0 = t(40 – 5t)
t = 0 or 40 – 5t = 0
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit
the ground?
d = rt – 5t2
0 = 40t – 5t2
0 = t(40 – 5t)
t = 0 or 40 – 5t = 0
40 = 5t
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit
the ground?
d = rt – 5t2
0 = 40t – 5t2
0 = t(40 – 5t)
t = 0 or 40 – 5t = 0
40 = 5t
8=t
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit
the ground?
d = rt – 5t2
0 = 40t – 5t2
0 = t(40 – 5t)
t = 0 or 40 – 5t = 0
40 = 5t
8=t
It will take 8 sec.
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How high will the ball be after 6.5
seconds?
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How high will the ball be after 6.5
seconds?
d = rt – 5t2
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How high will the ball be after 6.5
seconds?
d = rt – 5t2
d = 40(6.5) – 5(6.5)2
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How high will the ball be after 6.5
seconds?
d = rt – 5t2
d = 40(6.5) – 5(6.5)2
d = 260 – 5(42.25)
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How high will the ball be after 6.5
seconds?
d = rt – 5t2
d = 40(6.5) – 5(6.5)2
d = 260 – 5(42.25)
d = 260 – 211.25
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How high will the ball be after 6.5
seconds?
d = rt – 5t2
d = 40(6.5) – 5(6.5)2
d = 260 – 5(42.25)
d = 260 – 211.25
d = 48.75 m
80 m in 4 seconds
75 m
3 sec
75 m
5 sec
60 m
2 sec
60 m
6 sec
48.75 m 6.5 sec
35 m
1 sec
35 m
7 sec
on the ground in 8 sec
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for
the ball to be 100 m off the ground?
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for
the ball to be 100 m off the ground?
d = rt – 5t2
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for
the ball to be 100 m off the ground?
d = rt – 5t2
100 = 40t – 5t2
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for
the ball to be 100 m off the ground?
d = rt – 5t2
100 = 40t – 5t2
5t2 – 40t + 100 = 0
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for
the ball to be 100 m off the ground?
d = rt – 5t2
100 = 40t – 5t2
5t2 – 40t + 100 = 0
5(t2 – 8t + 20) = 0
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for
the ball to be 100 m off the ground?
d = rt – 5t
2
100 = 40t – 5t2
5t2 – 40t + 100 = 0
5(t2 – 8t + 20) = 0
t2 – 8t + 20 = 0
d = rt – 5t2
A person standing on the ground
throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for
the ball to be 100 m off the ground?
d = rt – 5t
2
100 = 40t – 5t2
5t2 – 40t + 100 = 0
5(t2 – 8t + 20) = 0
t2 – 8t + 20 = 0
The discriminant is negative so
there is no answer. In this case that
means that the ball never reaches
100 m off the ground.
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
0-meters
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
0-meters
-10-meters
d = rt – 5t2
-10 = r • 1.9 – 5(1.9)2
d = rt – 5t2
-10 = r • 1.9 – 5(1.9)2
-10 = 1.9r – 5 • 3.61
d = rt – 5t2
-10 = r • 1.9 – 5(1.9)2
-10 = 1.9r – 5 • 3.61
-10 = 1.9r – 18.05
d = rt – 5t2
-10 = r • 1.9 – 5(1.9)2
-10 = 1.9r – 5 • 3.61
-10 = 1.9r – 18.05
8.05 = 1.9r
d = rt – 5t2
-10 = r • 1.9 – 5(1.9)2
-10 = 1.9r – 5 • 3.61
-10 = 1.9r – 18.05
8.05 = 1.9r
4.24 r
d = rt – 5t2
-10 = r • 1.9 – 5(1.9)2
-10 = 1.9r – 5 • 3.61
-10 = 1.9r – 18.05
8.05 = 1.9r
4.24 r
Her initial upward velocity
was about 4.24 m/sec.
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
How long long will it take for the diver to be
even with the platform on the way down?
0 = 4.24t – 5t2
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
How long long will it take for the diver to be
even with the platform on the way down?
0 = 4.24t – 5t2
0 = t(4.24 – 5t)
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
How long long will it take for the diver to be
even with the platform on the way down?
0 = 4.24t – 5t2
0 = t(4.24 – 5t)
0=t
0 = 4.24 – 5t
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
How long long will it take for the diver to be
even with the platform on the way down?
0 = 4.24t – 5t2
0 = t(4.24 – 5t)
0=t
0 = 4.24 – 5t
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
How long long will it take for the diver to be
even with the platform on the way down?
0 = 4.24t – 5t2
0 = t(4.24 – 5t)
0=t
0 = 4.24 – 5t
5t = 4.24
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
How long long will it take for the diver to be
even with the platform on the way down?
0 = 4.24t – 5t2
0 = t(4.24 – 5t)
0=t
0 = 4.24 – 5t
5t = 4.24
t = 0.848 sec.
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
What is the highest point the diver will get from
the surface of the water?
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
What is the highest point the diver will get from
the surface of the water?
0-meters
-10-meters
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
What is the highest point the diver will get from
the surface of the water?
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
What is the highest point the diver will get from
the surface of the water?
0.848 ÷ 2 = 0.424 sec.
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
What is the highest point the diver will get from
the surface of the water?
0.848 ÷ 2 = 0.424 sec.
d = 4.24(0.424) – 5(0.424)2
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
What is the highest point the diver will get from
the surface of the water?
0.848 ÷ 2 = 0.424 sec.
d = 4.24(0.424) – 5(0.424)2
d = 1.798 – 5(0.180)
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
What is the highest point the diver will get from
the surface of the water?
0.848 ÷ 2 = 0.424 sec.
d = 4.24(0.424) – 5(0.424)2
d = 1.798 – 5(0.180)
d = 1.798 – 0.899
A diver dives from a 10-meter platform by
jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into
the water, what was her initial upward velocity?
What is the highest point the diver will get from
the surface of the water?
0.848 ÷ 2 = 0.424 sec.
d = 4.24(0.424) – 5(0.424)2
d = 1.798 – 5(0.180)
d = 1.798 – 0.899
d = 0.899 m or about 0.9 m
The diver gets about
0.9 meters above the platform
so she is 10.9 meters above
the water.
The formula, d = rt – 5t2, is often
used when working with vertical
motion. The rate (r) is m/sec., so the
distance (d) is in meters and the time
(t) is in seconds. The formula is a
quadratic equation. A related formula
is d = rt – 16t2, where the rate is
given in ft/sec.