Spark image

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 (vo) = 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 + (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.)
 

A VERSION IN WORD IS AVAILABLE ON THE SCHOOLPHYSICS USB
 
 
 
 
© Keith Gibbs 2020
 

A VERSION IN WORD IS AVAILABLE ON THE SCHOOLPHYSICS USB
 
 
 
 
© Keith Gibbs 2020