projectile motion
DESCRIPTION
Powerpoint on How to Solve Projectile Motion ProblemsTRANSCRIPT
![Page 1: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/1.jpg)
Projectile Motion
![Page 2: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/2.jpg)
What is a projectile motion?
![Page 3: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/3.jpg)
A projectile motion is a two-dimensional
motion with the force of gravity acting upon
it.
![Page 4: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/4.jpg)
TERMS
![Page 5: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/5.jpg)
Projectile
It is an object thrown by the exertion of a force, having its
own trajectory.
![Page 6: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/6.jpg)
Trajectory
It is the curved path taken by a projectile.
![Page 7: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/7.jpg)
Time of Flight
It is the duration in which the projectile is in flight.
![Page 8: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/8.jpg)
Range It is the horizontal
displacement from the takeoff point to the landing
point.
![Page 9: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/9.jpg)
Ballistics
It is the study of projectile motion.
![Page 10: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/10.jpg)
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.
![Page 11: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/11.jpg)
Projections
![Page 12: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/12.jpg)
Horizontal Projection
![Page 13: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/13.jpg)
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.
![Page 14: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/14.jpg)
The Three Kinematic Equations
![Page 15: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/15.jpg)
D = displacementa = acceleration
t = timevf = final velocityvi = initial velocity
![Page 16: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/16.jpg)
Solving the Horizontal Projection
![Page 17: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/17.jpg)
Determine the horizontal and vertical components
of the problem.
![Page 18: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/18.jpg)
HorizontalX = horizontal displacement
Ax = horizontal acceleration (0 m/s2)
T = timeVfx = final horizontal velocity
Vix = initial horizontal velocity
![Page 19: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/19.jpg)
Vertical
y = vertical displacementAy = vertical acceleration (-9.8
m/s2) T = time
Vfy = final vertical velocity
Viy = initial vertical velocity (0 m/s2)
![Page 20: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/20.jpg)
Draw the trajectory.
![Page 21: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/21.jpg)
Identify the unknown.
![Page 22: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/22.jpg)
Solve the missing by manipulating the three kinematic equations.
![Page 23: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/23.jpg)
Horizontal
![Page 24: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/24.jpg)
Vertical
![Page 25: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/25.jpg)
Other Equations
![Page 26: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/26.jpg)
Trajectory
𝑦=12𝑔 [ 𝑥𝑣 𝑖 ]
2
![Page 27: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/27.jpg)
Velocity at any instant of time
![Page 28: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/28.jpg)
Time of Flight
𝑡=√ 2 𝑦𝑔
![Page 29: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/29.jpg)
Range
𝑟=𝑣 𝑖𝑥 √ 2 𝑦𝑔
![Page 30: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/30.jpg)
Sample Problems
![Page 31: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/31.jpg)
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.
![Page 32: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/32.jpg)
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?
![Page 33: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/33.jpg)
Solutions
![Page 34: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/34.jpg)
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 = ?
![Page 35: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/35.jpg)
Find t.
![Page 36: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/36.jpg)
Solve for vix
![Page 37: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/37.jpg)
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 = ?
![Page 38: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/38.jpg)
Find t.
![Page 39: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/39.jpg)
Solve for x
![Page 40: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/40.jpg)
Angular Projection
![Page 41: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/41.jpg)
In the angular projection, the projectile encounters
an upward motion, therefore the maximum height is needed to be
solved.
![Page 42: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/42.jpg)
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.
![Page 43: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/43.jpg)
Solving the Angular
Projection
![Page 44: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/44.jpg)
Horizontal Initial Velocity
𝑣 𝑖𝑥=𝑣 𝑖cos𝜃
![Page 45: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/45.jpg)
Vertical Initial Velocity
𝑣 𝑖𝑦=𝑣 𝑖 sin𝜃
![Page 46: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/46.jpg)
Other Equations
![Page 47: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/47.jpg)
Velocity at any instant of time
![Page 48: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/48.jpg)
Time of Flight
𝑇=2 (𝑣 𝑖𝑦 )𝑔
![Page 49: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/49.jpg)
Range
𝑟=𝑣 𝑖𝑥 √ 2 𝑦𝑔
![Page 50: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/50.jpg)
Maximum Height
𝐻=𝑣 𝑖𝑦𝑇2−12𝑔(𝑇2 )
2
![Page 51: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/51.jpg)
Sample Problems
![Page 52: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/52.jpg)
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.
![Page 53: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/53.jpg)
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.
![Page 54: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/54.jpg)
Solutions
![Page 55: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/55.jpg)
Given
Horizontal Vertical
x = ? y =
vix = ? viy = ?
Vfx = ? Vfy = ?
ax = 0 m/s2 ay = -9.8 m/s2
t = ? t = ?
![Page 56: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/56.jpg)
Find vix
![Page 57: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/57.jpg)
Find viy
![Page 58: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/58.jpg)
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 = ?
![Page 59: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/59.jpg)
Find t.
![Page 60: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/60.jpg)
Solve for x
![Page 61: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/61.jpg)
Solve for y
![Page 62: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/62.jpg)
Given
Horizontal Vertical
x = ? y = ?
vix = ? viy = ?
Vfx = ? Vfy = ?
ax = 0 m/s2 ay = -9.8 m/s2
t = ? t = ?
![Page 63: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/63.jpg)
Find vix
![Page 64: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/64.jpg)
Find viy
![Page 65: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/65.jpg)
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 = ?
![Page 66: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/66.jpg)
Find t from the launch to the maximum height
1
0 1
1
![Page 67: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/67.jpg)
Solve for y
![Page 68: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/68.jpg)
Find the freefall t2
16. 4
![Page 69: Projectile Motion](https://reader038.vdocuments.mx/reader038/viewer/2022103022/55cf921f550346f57b93d96f/html5/thumbnails/69.jpg)
Solve for x