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## King Fahd University of Petroleum

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### King Fahd University of Petroleum & Minerals Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 4 12.6 Motion of a Projectile Projectile ... – PowerPoint PPT presentation

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Title: King Fahd University of Petroleum

1
King Fahd University of Petroleum Minerals
• Mechanical Engineering
• Dynamics ME 201
• BY
• Lecture 4

2
12.6 Motion of a Projectile
• Projectile any body that is given an initial
velocity and then follows a path determined by
the effects of gravitational acceleration and air
resistance.
• Trajectory path followed by a projectile

3
Horizontal Motion is Uniform Motion
Notice that the Horizontal motion is in no way
affected by the Vertical motion.
4
Projectile
5
Horizontal and vertical components of
velocity are independent.
Vertical velocity decreases at a constant
rate due to the influence of gravity.
6
Verify this mathematically
7
Horizontal Motion
• Acceleration ax 0
• Conclusion 1 Horizontal velocity remains
constant
• Conclusion 2 Equal distance covered in equal
time intervals

8
Vertical Motion
• ac -g 9.81 m/s2 32.2 ft/s2
• Conclusion 1 Equal increments of speed gained
in equal increments of time
• Distance increases in each time interval

9
Projectile Motion
• Assumptions
• (1) free-fall acceleration
• (2) neglect air resistance
• Choosing the y direction as positive upward
• ax 0 ay - g (a constant)
• Take x0 y0 0 at t 0
• Initial velocity v0 makes an
• angle ?0 with the horizontal

10
Maximum Height
At the peak of its trajectory, vy
0. From Time t1 to reach the peak Substituting
into
11
Projection Angle
• The optimal angle of projection is dependent on
the goal of the activity.
• For maximal height the optimal angle is 90o.
• For maximal distance the optimal angle is 45o.

12
Projection angle 10 degrees
10 degrees
13
Projection angle 45 degrees
10 degrees 30 degrees 40 degrees 45 degrees
14
Projection angle 60 degrees
10 degrees 30 degrees 40 degrees 45 degrees 60
degrees
15
Projection angle 75 degrees
10 degrees 30 degrees 40 degrees 45 degrees 60
degrees 75 degrees
So angle that maximizes Range (qoptimal) 45
degrees
16
Example
• A ball is given an initial velocity of V0 37
m/s at an angle of q 53.1. Find the position
of the ball, and the magnitude and direction of
its velocity, when t 2.00 s. Find the time when
the ball reaches the highest point of its flight,
and find its height h at this point
• The initial velocity of the ball has components
• v0x v0 cos ?0 (37.0 m/s) cos 53.1 22.2
m/s
• v0y v0 sin ?0 (37.0 m/s) sin 53.1 29.6 m/s
• position
• x v0xt (22.2 m/s)(2.00 s) 44.4 m
• y v0yt - ½gt2
• (29.6 m/s)(2.00 s) ½ (9.80 m/s2)(2.00 s)2
• 39.6 m

17
Solution (con.)
• Velocity
• vx v0x 22.2 m/s
• vy v0y gt 29.6 m/s (9.80 m/s2)(2.00 s)
10.0 m/s

18
Solution (cont.)
• b) Find the time when the ball reaches the
highest point of its flight, and find its height
H at this point.

19
Solution (cont.)
c) Find the horizontal range R, (that is, the
horizontal distance from the starting point to
the point at which the ball hits the ground.)
20
A ball traveling at 25 m/s drive off of the edge
of a cliff 50 m high. Where do they land?
Horizontally x x0 (v0)x t
x 25 3.19 79.8 m
Initial Conditions vx 25 m/s vy0 0 m/s a
- 9.8 m/s2 t 0 y0 0 m y - 50 m x0 0 m
Vertically v v0-gt y y0 v0t 1/2gt2 . v2
v02 - 2g(y-y0).
-50 001/2(-9.8)t2 t 3.19 s
21
Review
• Example 12.11
• Example 12.12
• Example 12.13

22
Thank you