Projectile Motion

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PROJECTILES
3 Types: Vertical, Horizontal, Inclined
An object is launched with
an initial velocity in the
vertical direction.
v = 0 m/s
At the peak of the object’s
flight, v = 0 m/s.
Vertical
Once the object returns to
the same height it was
released from, its velocity
will be the same as its
initial.
vi
Object starts out with an
initial velocity in the
horizontal direction.
Acceleration due to
gravity gives the object a
velocity in the vertical
direction.
Horizontal
An object is launched with a velocity that has vertical and horizontal
components. These components can be resolved by setting up a right
triangle.
vi
Inclined
Independence of Motion
The vertical and horizontal motion (free fall) does not affect
the horizontal motion (constant velocity)
The Horizontal Projectile
vi
In order to find the
vertical distance
travelled, we have to
separate the horizontal
motion from the
vertical motion.
dy
Vertical Motion (y)
viy = 0 m/s
ay = -9.8 m/s2
dy = height
to find the time, use:
dy = viyt + ½ayt2
dy = ½ayt2
Horizontal Motion (x)
vix = vi
ax = 0 m/s2
dx = vixt + ½axt2
dx = vixt
Note: - ax = 0!!!! Gravity affects only vertical motion.
- time of flight is the same regardless of vi, a
dropped bullet and a fired bullet hit the ground at
the same time.
Practice Problem 1
A pool ball leaves a 0.60-meter high table
with an initial horizontal velocity of
2.4 m/s. Predict the time required for the
pool ball to fall to the ground and the
horizontal distance between the table's
edge and the ball's landing location.
Practice Problem 2
A soccer ball is kicked horizontally off a
22.0-meter high hill and lands a distance
of 35.0 meters from the edge of the hill.
Determine the initial horizontal velocity of
the soccer ball.
TRAJECTORY – THE PATH OF A PROJECTILE
For Horizontal projectile motion:
-ax = 0 m/s2! Gravity affects only vertical motion
-time of flight is the same regardless of v i, a dropped bullet and a fired bullet hit
the ground at the same time.
Horizontal Motion
Mythbusters Bullet Drop
THE INCLINED PROJECTILE
Vy = 0 m/s at top
viy
vi
θ
vix
Initial velocity
must be resolved
into components
(viy and vix)
HOW DO WE SOLVE THIS??
Vertical Motion (y)
viy = visinθ very important!!
ay = -9.81 m/s2
vfy = 0 m/s at top
to find time:
use vfy = viy + ayt to get time to
top, then double it.
to find max height:
vfy2 = viy2 + 2aydy
Horizontal Motion (x)
Vix = vicosθ
ax = 0 m/s2
t obtained from vertical to
find range:
dx = vixt + ½ axt2
0
PRACTICE PROBLEM #1
An airplane traveling 1001 m above the ocean at 125
km/h is going to drop a box of supplies to
shipwrecked victims below.
a. How many seconds before the plane is
directly overhead should the box be dropped?
b. What is the horizontal distance between the
plane and the victims when the box is dropped?
PRACTICE ANSWER #1
Vertical
viy = 0 m/s
ay = -9.8 m/s2
dy = 1001 m
t=?
d = ½ at2
1001 m = ½ (9.8 m/s2)t2
t = 14.3 s
Horizontal
vix = 125 km/h
dx = ?
t=?
ax = 0 m/s2
dx = vixt
dx = (125 km/h)(14.3 s)
dx = (0.03472 km/s)(14.3 s)
dx = 0.4965 km = 496.5 m
PRACTICE PROBLEM #2
Herman the human cannonball is launched
from level ground at an angle of 30° above
the horizontal with an initial velocity of 26
m/s. How far does Herman travel
horizontally before reuniting with the
ground?
PRACTICE ANSWER #2
Vertical
viy = 26(sin30)
ay = -9.8 m/s2
vf = 0 m/s
t=?
vf = viy + at
0 = 26(sin30) + (-9.8 m/s2)t
-13 = -9.8 m/s2t
t = 1.33 s
Horizontal
vix = 26(cos30)
ax = 0 m/s2
dx = ?
t=?
dx = vixt
dx = 26(cos30)(2.66 s)
dx = 60.0 m
EFFECT OF LAUNCH ANGLE
1. Complementary angles produce the same range.
2. Maximum range is at 45 degrees launch angle.
3. Higher angle leads to longer “hang time”
75 °
60 °
30 °
15°
Launch
45 °
TRAJECTORY
vy = 0
vx
vy
vx
vx
viy
vy
vx
vix
vy

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