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Transcript
```Projectile motion and drag
Question:
A 0.142 kg baseball is dropped from rest and has a terminal speed of 42.5 m/s (95 mph).
(a) If a baseball experiences a drag force of magnitude r = Cv2, what is the value of the
constant?
(b) What is the magnitude of the drag force when the speed of the baseball is 36 m/s?
(c) Use a computer to determine the motion of a baseball thrown vertically upward at an
initial speed of 36.0 m/s.
(i) What maximum height does the ball reach?
(ii) What is its speed just before it hits the ground?
When the ball reaches its terminal speed there will be no vertical acceleration and so the
resistive or drag force will be equal to the weight of the baseball.
Therefore:
Drag force = r = mg = 0.142x9.81 = 1.393 = Cv2 = Cx42.52 = Cx1806.25
C = 1.393/1806.25 = 0.00077 Ns2m-2.
(b) when the speed = 36 ms-1 the drag force (r) is given by the equation:
r = Cv2 = 0.00077 x 362 = 0.999 = 1 N
(c) This problem can be solved by an application of Euler's method using a spreadsheet. The
following version shows an extract from such a spreadsheet.
Motion of body projected upwards and affected by air resistance
This section shows the use of a spreadsheet to calculate the properties of the motion of an
object projected vertically upwards and affected by air resistance.
Velocity at any point = v
Drag force is proportional to v2
Drag force = Cv2 where C is the drag coefficient [= 0.00077 in this case]
Acceleration due to gravity = g = 9.81 ms-2
Initial vertical velocity (vo) = 36 ms-1
Values calculated at time interval t = 0.05 s
Mass of object (m) = 0.142 kg
Equations:
New vertical velocity (v') = v - gt – bv2 where b = Cm
(v is the previous vertical velocity)
New vertical height (y') = y + ½ (v + v')t
(y is the previous vertical height)
The following table shows the values for t,v and y. Not all values of t are calculated.
Time (t) (s) Vertical velocity (ms-1) Vertical height (m)
0
36
0
0.05
35.16
1.78
0.1
34.35
3.52
0.15
33.54
5.21
0.5
28.29
16.02
1.0
21.68
28.47
1.5
15.81
37.82
2.0
10.43
44.36
2.5
5.35
48.30
3.0
0.42
49.74
3.04
0.03
49.75
3.2
1.54
49.63
3.5
4.47
48.73
4.0
9.24
45.29
4.2
11.09
43.26
4.5
13.79
39.52
5.0
18.0
31.58
6.0
25.23
9.80
6.1
25.86
7.25
6.30
27.06
1.95
6.37
27.5
0.04
The program does not give us the exact velocity at y' = 0 since the smallest value given for y'
is 0.04 m.
The velocity on reaching the ground is therefore slightly greater than 27.5 ms-1 and the total
time of flight is a little over 6.37 s.
The maximum height reached is 49.75 m after about 3.04 s.
Useful websites for the further analysis of this problem.
http://www.engapplets.vt.edu/dynamics/projectile/projectile6/Projectile.html