THESE
TIMES ARE
EQUAL
The time it takes for
that object to hit the
ground if I just drop
it.
viy = 0
The time it takes for
an object thrown
horizontally to hit the
ground
Look how gravity
affects vertical motion
… the distance gets
bigger.
Horizontal velocity is
constant …
Always use this
formula
For the horizontal part, gravity
plays no role. The distance
between points is always the same.
WORKSHEET 1: #1
Tom throws a rock horizontally off of a
40.0 m high cliff. How fast did he throw
the rock if it hits the ground 45.0 m from
the base of the cliff?
It takes the same amount of time to hit the
ground if I throw it horizontally as it
does if I simply drop it … i.e., vi = 0
I need vx. Find an equation
with this, and see what I’m
missing.
d y  v iy t 
dy 
vx 
d x 45.0

t
t
t
2d y
1
1
gt
2
2
gt
2
2

g
2  40 
 2.86 s
9.8
I need
t
vx 
45
2.86
 15.7 m / s
Each dot = ½ second
WORKSHEET 1: #2
Erica throws a ball out of a window
with a horizontal speed of 3.80 m/s. If
the ball hit the ground 15.0 m away,
how high up is the window?
It takes the same amount of time to hit the
ground if I throw it horizontally as it
does if I simply drop it … i.e., vi = 0
I need dy. Find an equation
with this, and see what I’m
missing.
1
d y  viyt  gt 2
2
dy 
1 2 1
gt   9.8 t 2
2
2
vx 
t
dx
t
d x 15.0

 3.95s
vx 3.80
I need
t
dy 
1
2
 9.81 3.95  76.5m
2
Each dot = ½ second
WORKSHEET 1: #3
Kelly shoots a rock in a sling shot
horizontally at 12.5 m/s from the top of a
30.0 m high building. How far from the
base of the building does the rock hit?
Each dot = ½ second
WORKSHEET 1: #4
Thelma drives her car off a cliff 60.0 m
high. It hits the ground 125 m out
horizontally from the bottom of the cliff.
How fast was she driving?
Each dot = ½ second
WORKSHEET 1: #5
Elaine shoots a bullet straight up in the air
from a gun. It reaches a height of 1940 m.
How fast was the bullet fired?
I need viy. Find an equation
with this, and see what I’m
missing.
At the highest point, the object
momentarily stops, i.e., vf = 0
v2fy  viy2  2 gd
0  viy2  2 gd
viy  2 gd

 2 9.811940 
 195m / s
Each dot = 1 second
WORKSHEET 1: #6
How long is the bullet in #5 in the air
before it hits the ground again?
I need t. Find an equation
with this, and see what I’m
missing.
t

2viy
g
 2 195
19.9 seconds (red dots)
going up, and 19.0
seconds coming down.
9.81
 39.8s
Each dot = 1 second
WORKSHEET 1: #7
If Elaine had shot the bullet from #5
horizontally from a height of 2.50 m, how
far would the bullet have gone before
hitting the ground?
Each dot = 0.1 seconds
WORKSHEET 2: #1
Chris throws a rock at 21.0 m/s into the air at 40.0°
above the horizontal. How long is the rock in the air?
NOTE:
With projectiles above the horizontal, I would
calculate both vectors first, and label them.
vx  21cos  40   16.1m / s
v y  21sin  40   13.5m / s
I need t. Find an equation
with this, and see what I’m
missing.
t

2viy
g
 2 13.5
9.81
 2.76s
Each dot = 0.25 seconds
WORKSHEET 2: #2
What is the maximum height for the rock
in #1?
I need d. Find an equation
with this, and see what I’m
missing.
v2fy  viy2  2 gd
0  viy2  2 gd
d
viy2
2g
 13.5

 2  9.81
2
At the highest point, the
velocity in the y direction = 0;
vyf = 0
 9.30m
Each dot = 0.25 seconds
WORKSHEET 2: #3
How far away did the rock in #1 land?
I need dx. Find an equation
with this, and see what I’m
missing.
vx 
dx
t
d x  vx t
 16.1 2.76s   44.4m
Each dot = 0.25 seconds
WORKSHEET 2: #4
Nicole throws a ball at 25.0 m/s at an
angle of 63.8° above the horizontal. What
was the range of the ball?
Each dot = 0.25 seconds
WORKSHEET 2: #5
Jason throws a ball at 25.0 m/s at an angle
of 26.2º above the horizontal. What is the
range of the ball?
Each dot = 0.25 seconds
WORKSHEET 2: #6
An airplane flying at 450 km/hr and 3330 m
above the ground drops a bomb on a target.
How far in front of the target will the plane
have to release the bomb in order to hit it?
Each dot = 1 second
WORKSHEET 2: #7
Alex throws a ball at 22.5 m/s at an angle of 34.8°
below the horizontal from the top of a 40.0 m high
building. How far from the base of the building
does the ball hit?
NOTE:
With projectiles above the horizontal, I would
calculate both vectors first, and label them.
vx  22.5cos  34.8  18.5m / s
v y  22.5sin  34.8  12.8m / s
I need dx. Find an equation
with this, and see what I’m
missing.
1
d y  viyt  gt 2
2
40  12.8t  .5 9.8 t 2
vx 
dx
t
4.9t 2  12.8t  40  0
d x  v x t   1 8 .5  t
t  1.84s
I need
t
d x  vxt  18.51.84  34.0m
Each dot = 0.25 seconds
WORKSHEET 2: #8
What is the maximum range for a
projectile fired with a velocity of 80 m/s?
WORKSHEET 3: #1
A ball is thrown at 33.3 m/s at an angle of 75.5º
from the ground. How high off the ground is the
ball after 6.00 s?
NOTE:
With projectiles above the horizontal, I would
calculate both vectors first, and label them.
vx  33.3cos  75.5  8.3m / s
v y  33.3sin  75.5  32.2m / s
I need dy. Find an equation
with this, and see what I’m
missing.
1
d y  viyt  gt 2
2
  32.2  6  
1
2
 9.81 6 
2
if we’re calling
velocity UP as
positive;
Then gravity,
pulling DOWN,
must be negative.
 16.8m
Each dot = 1 second
WORKSHEET 3: #2
A kicker wants to kick a 45.7 m (50 yd) field goal.
The crossbar of the upright is 3.05 m above the
ground. The kicker can kick with a velocity of
22.4 m/s. The ball is kicked at an angle of 50.0°
above the ground. If the ball is kicked straight, is
the field goal good?
vx  22.4cos  50   14.4m / s
v y  22.4sin  50   17.2m / s
I need dy. Find an equation
with this, and see what I’m
missing.
1
d y  viyt  gt 2
2
d y  17.2  t  
1
2
 9.81 t 
2
vx 
dx
t
t
d x 45.7

 3.17 s
vx 14.4
I need
t
d y  17.2  3.17  
1
2
 9.81 3.17   5.28m
2
Each dot = 0.1 seconds
WORKSHEET 3: #3
A ball is kicked at 30.0 m/s at an angle of 65.0º
above the ground. How long has the ball been in the
air when the velocity of the ball is pointing down
and has a magnitude of 15.0 m/s?
Remember .. In the horizontal
direction, velocity is constant
I need t. Find an equation
with t in it, and see what I’m
missing.
?
vf
15
a
v

v f  vi
t
t 
t
v f  1 2 .7  1 5
2
v f  vi
2
2
g
v f  225  161.29  63.71
65
vx  12.7
v y  27.2
v
30
2
vx  12.7
I need
vf
t
v f  27.2
v f  7.98
 9.8
(Since velocity is going down in
direction, it is negative.)
t
 7.98  27.2
 9.8
 3.59 s
t = 3.59 seconds …
We are right!
WORKSHEET 3: #4
A bullet is fired at 200. m/s at an angle of
56.7°. At what angle is the bullet moving
when it is 100. m above the ground?
tan    
v yf
v yf  ?
vyf2  vyi2  2ad
110

110
v 
   tan 1  yf 
 110 
vyf 
If I knew vyf, I
could find the
angle by using
inverse tangent.
167    2 9.8100
2
 161
I need
Remember .. In the
horizontal direction,
velocity is constant
vf
 161 
  55.7
110


   tan 1 
200
167
56.7°
110
Each dot = 0.1 seconds
WORKSHEET 3: #5
Jane is standing on top of a 50.0 m high building. She
throws a ball at 20.0 m/s at an angle of 30.0º above
the horizontal. If she was standing on the edge of the
building when she threw the ball, how far from the
base of the building does the ball hit the ground?
I need dx. Find an
equation with in it, and
see what I’m missing.
d
vx  x
t
d x  vx t
1
d y  viyt    at 2
2
50  10t  .5 9.8 t 2
4.9t 2  10t  50  0 Use quadratic formula
to solve for t.
t  4.37s
I need
t
d x   1 7 .3   4 .3 7   7 5 .6 m
Each dot = 0.25 seconds