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 = Cv^{2}, 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 = Cv^{2} = Cx42.5^{2} = Cx1806.25

C = 1.393/1806.25 = 0.00077
Ns^{2}m^{-2}.

(b) when the speed = 36 ms-1 the drag force (r) is given
by the equation:

r = Cv^{2} = 0.00077 x 36^{2} = 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.

Velocity at any point = v

Drag force is proportional to v

Drag force = Cv

Acceleration due to gravity = g = 9.81 ms

Initial vertical velocity (v

Values calculated at time interval Δt = 0.05 s

Mass of object (m) = 0.142 kg

Equations:

New vertical velocity (v') = v – gDt – bv

(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

The maximum height reached is 49.75 m after about 3.04 s.

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.)