Answer:
drag force [tex]F_D = 1.5 \ N[/tex]
Velocity (V) = 40.169 m/s
Explanation:
The drag force [tex]F_D[/tex] is given by the formula:
[tex]F_D = C_D * \frac{1}{2}* \rho * V^2*A[/tex]
where:
[tex]C_D[/tex] = drag coefficient depending on the Reynolds number
Reynolds number Re = [tex]\frac{\rho *V*D}{ \mu}[/tex]
Let's Assume that the air is in room temperature at 25 °C ; Then
density of the air [tex]\rho[/tex] = 1.1845 kg/m³
viscosity of fluid or air [tex]\mu[/tex] = 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 :
[tex]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[/tex]
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:
[tex]C_D[/tex] = 0.3
Projected Area A = [tex]\frac{\pi D^2}{4}[/tex]
A = [tex]\frac{\pi 0.074^2}{4}[/tex]
A = 0.0043 m²
Finally, drag force is then calculated as ;
[tex]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[/tex]
b)
[tex]- F_D = ma[/tex]
since acceleration a = [tex]\frac{dV}{dt}[/tex]
Then;
[tex]-F_D = m \frac{dV}{dt}[/tex]
Also;
velocity (V) = [tex]\frac{dx}{dt}[/tex]
Then;
[tex]- F_D = \frac{md_2x}{dt^2}[/tex]
[tex]\frac{d_2x}{dt^2} = \frac {- F_D}{m}[/tex]
[tex]F_D = 1.5 \ N\\m = 0.15 \ kg[/tex]
Then;
[tex]\frac{d_2x}{dt^2} = \frac {- 1.5 }{0.15}[/tex]
[tex]\frac{d_2x}{dt^2} =- 10[/tex]
Integrating the above equation ; we have :
[tex]\frac{dx}{dt}= - 10 t + C\\[/tex]
when time (t) = 0 ; then [tex]\frac{dx}{dt}= V = 44.3[/tex]
44.3 = - 10 × 0 + C
C = 44.3
[tex]\frac{dx}{dt}= V = -10 t + 44.3[/tex]
Time (t) =
[tex]\frac{distance }{velocity} \\\\= \frac{18.3 m}{44.3 m/s}\\\\= 0.413 s[/tex]
∴ Velocity ; [tex]\frac{dx}{dt}= V = - 10t +44.3[/tex]
[tex]\frac{dx}{dt}= V = - 10(0.413 s) +44.3[/tex]
Velocity (V) = 40.169 m/s
