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?

Answer:

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 (v_o) = 36 ms^-1
Values calculated at time interval Δt = 0.05 s
Mass of object (m) = 0.142 kg

Equations:
New vertical velocity (v') = v – gDt – bv² where b = Cm
(v is the previous vertical velocity)
New vertical height (y') = y + ½ (v + v')Δt
(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

http://math.fullerton.edu/mathews/n2003/projectilemotion/ProjectileMotionMod/ Links/ProjectileMotionMod_lnk_4.html

http://galileo.phys.virginia.edu/classes/581/Pr ojectilesExcel.html

(If you use this last link do not forget to modify the equations to allow for the mass of the baseball.)