projectile motion
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Projectile Motion
What is a projectile motion?
A projectile motion is a two-dimensional
motion with the force of gravity acting upon
it.
TERMS
Projectile
It is an object thrown by the exertion of a force, having its
own trajectory.
Trajectory
It is the curved path taken by a projectile.
Time of Flight
It is the duration in which the projectile is in flight.
Range It is the horizontal
displacement from the takeoff point to the landing
point.
Ballistics
It is the study of projectile motion.
Since projectile motion is a two-dimensional motion, it
has a horizontal and vertical motion. The horizontal
motion is constant, while the vertical motion is
constantly changing due to the force of gravity.
Projections
Horizontal Projection
In the horizontal projection, the projectile
doesn’t encounter an upward motion. It is directed horizontally
straight and goes downward until it reaches
a surface.
The Three Kinematic Equations
D = displacementa = acceleration
t = timevf = final velocityvi = initial velocity
Solving the Horizontal Projection
Determine the horizontal and vertical components
of the problem.
HorizontalX = horizontal displacement
Ax = horizontal acceleration (0 m/s2)
T = timeVfx = final horizontal velocity
Vix = initial horizontal velocity
Vertical
y = vertical displacementAy = vertical acceleration (-9.8
m/s2) T = time
Vfy = final vertical velocity
Viy = initial vertical velocity (0 m/s2)
Draw the trajectory.
Identify the unknown.
Solve the missing by manipulating the three kinematic equations.
Horizontal
Vertical
Other Equations
Trajectory
𝑦=12𝑔 [ 𝑥𝑣 𝑖 ]
2
Velocity at any instant of time
Time of Flight
𝑡=√ 2 𝑦𝑔
Range
𝑟=𝑣 𝑖𝑥 √ 2 𝑦𝑔
Sample Problems
A soccer ball is kicked horizontally off a 22-
meter high hill and lands a distance of 35 m from
the edge of the hill. Determine the initial
horizontal velocity of the soccer ball.
Fred throws a baseball 42 m/s horizontally from a
height of 2 m. How far will the ball travel before it reaches the ground?
Solutions
Given
Horizontal Vertical
x = 35 m y = -22 m
vix = ? viy = 0 m/s
Vfx = ? Vfy = ?
ax = 0 m/s2 ay = -9.8 m/s2
t = ? t = ?
Find t.
Solve for vix
Given
Horizontal Vertical
x = ? y = -2 m
vix = -42 m/s viy = 0 m/s
Vfx = -42 m/s Vfy = ?
ax = 0 m/s2 ay = -9.8 m/s2
t = ? t = ?
Find t.
Solve for x
Angular Projection
In the angular projection, the projectile encounters
an upward motion, therefore the maximum height is needed to be
solved.
The same kinematic equations are to be used in
solving the problem, but what differs is the initial
velocity of both horizontal and vertical, and also the maximum height of the
trajectory.
Solving the Angular
Projection
Horizontal Initial Velocity
𝑣 𝑖𝑥=𝑣 𝑖cos𝜃
Vertical Initial Velocity
𝑣 𝑖𝑦=𝑣 𝑖 sin𝜃
Other Equations
Velocity at any instant of time
Time of Flight
𝑇=2 (𝑣 𝑖𝑦 )𝑔
Range
𝑟=𝑣 𝑖𝑥 √ 2 𝑦𝑔
Maximum Height
𝐻=𝑣 𝑖𝑦𝑇2−12𝑔(𝑇2 )
2
Sample Problems
A long jumper leaves the ground with an initial
velocity of at an angle of above the ground.
Determine the time of flight, the horizontal distance, and the peak height of the long-
jumper.
A cannonball was shot 60 m from a cliff at an angle of . It has an initial velocity of . Find the maximum height,
time of flight and the horizontal displacement of
the cannonball.
Solutions
Given
Horizontal Vertical
x = ? y =
vix = ? viy = ?
Vfx = ? Vfy = ?
ax = 0 m/s2 ay = -9.8 m/s2
t = ? t = ?
Find vix
Find viy
Horizontal Vertical
x = ? y =
vix = 10.6 m/s viy = 5.63 m/s
Vfx = 10.6 m/s Vfy = -5.63 m/s
ax = 0 m/s2 ay = -9.8 m/s2
t = ? t = ?
Find t.
Solve for x
Solve for y
Given
Horizontal Vertical
x = ? y = ?
vix = ? viy = ?
Vfx = ? Vfy = ?
ax = 0 m/s2 ay = -9.8 m/s2
t = ? t = ?
Find vix
Find viy
Horizontal Vertical
x = ? y = ?
vix = 15.05 m/s viy = 19.97 m/s
Vfx = 15.05 m/s Vfy = ?
ax = 0 m/s2 ay = -9.8 m/s2
t = ? t = ?
Find t from the launch to the maximum height
1
0 1
1
Solve for y
Find the freefall t2
16. 4
Solve for x
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