A bullet is shot, a baseball is thrown, and a rock is
dropped –
which object will hit the ground first if released from the
top of the IDS Building?
Projectile Motion
Part 1
Projectiles launched horizontally
In order to answer this question, we must understand
projectile motion.
What is a projectile?
z
Any body that is thrown or projected into the air.
z
Projectiles follow a curved path near the earth’s surface
because of gravity’s effect on them.
Let’s start with the dropped rock.
z
The falling rock is only under the influence of gravity.
Therefore, if we know the height of the IDS building, we
can calculate the time in which it takes to fall.
Remember: d = 1/2gt2
z
With any curved path, we can break the motion down
into a vertical component and a horizontal component.
or
t = 2d
g
y
x
The baseball and the bullet can be treated
similarly to one another.
z
In the case of a projectile shot horizontally, it will have a
horizontal velocity vector and a vertical velocity vector.
IDS Center
However, the only force that pulls the ball or bullet to the ground is gravity!
So, which object will hit the ground first?
The rock, the baseball, and the bullet will all land at the
same time! (Assuming there is no air resistance)
Will the rock, the bullet and the baseball all land
in the same place?
NO!
Therefore, we need to look at that component separately.
Therefore, the bullet and the ball will accelerate toward the ground at the
same rate as the dropped rock.
1
How to determine the landing point of a projectile:
Let’s look at a similar question:
z
z
Where a projectile lands is determined by the
horizontal speed or velocity at which the object
is released.
Ignore the vertical pull of gravity at this time.
A green ball is dropped from a cliff 20 meters above the ground. At
the same time of release a second ball, red in color, is thrown
horizontally off the cliff at a speed of 35 m/s.
For example:
If a ball is thrown horizontally at a velocity of 20 m/s,
then after 1 second, it will traveled 20 meters, after 2 seconds,
40 meters, after 3 seconds, it has traveled 60 meters, etc…
Both balls will land at the same time.
The both are under the same influence of gravity.
20 m
40 m
Which ball will land first?
Where will each ball land in respect to the cliff?
60 m
That answer depends on the amount of time the balls are in the air.
The distance that the ball travels horizontally depends on the time
it is in the air.
To decide where the balls land, we need to determine how
z
The green ball will land directly under the point at which
it was dropped. It goes straight down.
z
But the red ball will land away from the cliff. The distance
that the red ball lands away from the cliff is determined
long the balls are in the air.
To do this, we need to use the equation that helps us
determine the amount of time that the balls are actually
falling. d = 1/2gt2
So, 20 m = ½ (9.8 m/s2) t 2
Solving for t, we get t =
20 m
or
(1/2)(9.8)m/s2
t=2s
by looking at only the horizontal component.
The red ball is traveling at 35 m/s at the release.
It travels in the air for 2 seconds.
v = 35 m/s
t = 2s
Using the equation, d = vt for linear motion,
The distance traveled in air is 35 m/s x 2 seconds = 70 meters.
Both balls will take 2 seconds to hit the ground.
The red ball will land 70 meters from the base of the cliff.
Let’s look at an example:
A stone is thrown horizontally off the top of a building 44 m
High with a speed of 15 m/s. How long will it take the stone
to reach the ground? How far away from the building will it
land?
15 m/s
Projectile Motion
Part 2
Projectiles Launched Upward
44 m
2
How does a projectile launched from the ground
differ from a free falling object?
Notice the arch of the ball thrown here:
Any projectile is a free falling object under
the influence of gravity.
z The only difference is that the projectile
has to go up, before it can come down.
z
The ball has a horizontal velocity component that does not change Similar to throwing the ball off the cliff.
It also has a vertical velocity vector that is under gravity’s effect.
This component is similar to one drawn if the ball was thrown
straight up
Determining the height and distance of a projectile:
z
z
Step 1: Find the component velocity vectors of
the take
- off velocity.
Step 2: Using the vertical velocity component,
determine the length of time the projectile is in
the air.
Find the time to reach a velocity of 0 m/s at the top.
Then, multiply by 2 for the complete path.
50
s
m/
x
y
Horizontal speed:
x = Cos Ө x 50 m/s
vtop = 0 m/s
Vertical speed:
y = Sin Ө x 50 m/s
vi
Determining the height of the projectile
Step 3: Using the time that the projectile takes to
reach the top of the path, calculate the height
(distance traveled upward) of the path.
Assume the projectile travels only vertically straight up or down.
Use the free-falling equation for distance.
dheight = 1/2gt2
vf - vi = gt
t = vf - vi
9.8 m/s
vf
Total time in air = t x 2
Determining the distance the projectile will travel:
Step 4: Using the horizontal velocity component,
determine how far the projectile will travel.
Assume the projectile travels only horizontally forward.
Use the linear motion equations.
d = vt (where t = the total time in the air )
3
Let’s look at an example:
A football is kicked from the ground to a position
downfield. If the ball is kicked with a speed of 20 m/s at a
45 degree angle, how long will it be in the air? How far
will the ball travel?
m
20
45°
/s
4