Question

The diameter of a baseball is 7.4 cm and its mass is 0.15 kg. a) If a pitcher throws the baseball at a velocity of 44.3 m/s (100 mi/hr), estimate the drag force on the baseball as it moves through the air. Assume that the surface of the baseball is smooth. Express your answer in units of Newtons. b) If the pitcher throws the baseball from the pitching mound at 44.3 m/s (100 mi/hr), estimate the baseball’s velocity when it crosses home plate, 18.3 m (60 ft) away. Express your answer in units of m/s. This is not steady state and therefore it may be useful to start from െF஽ ൌ ma and to use the relationship u ൌ dx d�

Answer 1

Answer:

drag force $F_D = 1.5 \ N$

Velocity (V) = 40.169 m/s

Explanation:

The drag force $F_D$ is given  by the formula:

$F_D = C_D * \frac{1}{2}* \rho * V^2*A$

where:

$C_D$ = drag coefficient depending on the Reynolds number

Reynolds number Re = $\frac{\rho *V*D}{ \mu}$

Let's Assume that the air is in room temperature at 25 °C ; Then

density of the air $\rho$ = 1.1845  kg/m³

viscosity of fluid or air $\mu$ = 1.844 × 10⁻⁵ kg/ms

diameter of the baseball D = 7.4 cm

Velocity V = 44.3 m/s

Replacing them into the equation of Reynolds number ; we have :

$Re = \frac{1.1845 \ kg/m^3*44.3 m/s*0.074 m}{1.844*10^{-5}kg/ms}\\\\Re = 2.1*10^5$

A = Projected Area

From the diagram attached below which is gotten from NASA for baseball;

the drag coefficient which depends on Reynolds number is read as:

$C_D$ = 0.3

Projected Area A = $\frac{\pi D^2}{4}$

A = $\frac{\pi 0.074^2}{4}$

A = 0.0043 m²

Finally, drag force is then calculated as ;

$F_D = C_D * \frac{1}{2}* \rho*V^2*A\\\\F_D = 0.3* \frac{1}{2}*1.1845 \ kg/m^3*(44.3 \ m/s) ^2*0.0043 m^2\\\\F_D = 1.5 \ N$

b)

$- F_D = ma$

since acceleration a = $\frac{dV}{dt}$

Then;

$-F_D = m \frac{dV}{dt}$

Also;

velocity (V) = $\frac{dx}{dt}$

Then;

$- F_D = \frac{md_2x}{dt^2}$

$\frac{d_2x}{dt^2} = \frac {- F_D}{m}$

$F_D = 1.5 \ N\\m = 0.15 \ kg$

Then;

$\frac{d_2x}{dt^2} = \frac {- 1.5 }{0.15}$

$\frac{d_2x}{dt^2} =- 10$

Integrating the above equation ; we have :

$\frac{dx}{dt}= - 10 t + C$

when time (t) = 0 ; then $\frac{dx}{dt}= V = 44.3$

44.3 = - 10 × 0 + C

C = 44.3

$\frac{dx}{dt}= V = -10 t + 44.3$

Time (t) =

$\frac{distance }{velocity} \\\\= \frac{18.3 m}{44.3 m/s}\\\\= 0.413 s$

∴ Velocity ; $\frac{dx}{dt}= V = - 10t +44.3$

$\frac{dx}{dt}= V = - 10(0.413 s) +44.3$

Velocity (V) = 40.169 m/s