Question

Last HW, we solved the problem of the flight of a softball assuming no drag from air resistance. Let’s add some complication to the problem by calculating the flight of the ball, allowing air resistance.

The formula for drag is: ?? = 1/2?? 2??? and always acts in the opposite direction to the flight of the ball Where: ?? = Force due to drag ? = density of fluid (air, assume at standard temperature and pressure) = .0765lb/ft3 ? = velocity ?? = Coefficient of drag; for a softball, this is somewhere in the range of 0.5 to 0.6 A = Cross sectional area of a softball. (Note Softballs are measured in circumference, the standard being 12 inches (who knew?)

And by the way, a standard softball must weigh between 6.25 and 7 ounces. Assume a softball player can throw a softball at 100ft/sec. Our objective is to calculate the optimal angle to throw the ball to get the runner out from 260 ft., now accounting for aerodynamic friction. We’ll get there, but first, we need to setup the problem for some arbitrary angle, say 30 degrees. How to go about solving this problem:

A. Draw a free-body diagram (Draw forces acting on this ball). Notice we have two forces acting on the ball now.

B. Write the component (x, and y) equations of motion. ?? = ? ?? ?? and ?? = ? ?? ?? Rewrite these equations, solving for ????? ??. [Note: You’ve been lied to your whole life if you learned that ? = ??. The equation is wrong. Actually it is, = ? ? ?? . The gravitational constant gc is required to keep the units of the equation straight.]

C. From the calculus course you took, ? = ∫?(?) ??. However, when you take the equations you developed in B, you cannot directly integrate them. (This is often the case.) Instead, you’ll have to integrate them numerically. What is done is to break up the function into small trapezoids and add them up. Shown as a diagram: Write formula for area of trapezoid from ti-1 to ti . The? ?? ≅ ??−1 + ???? ???? ??−1 ?? ?? Complete this equation for both components x and y.

D. Rewrite this equation as an equation in Excel

E. Redo steps C and D to calculate distance. (Hint: instead on integrating acceleration, you will integrate velocity.

F. Complete the Excel spreadsheet. (Remember, the y component of acceleration due to drag will change sign when vy becomes negative. You can have Excel handle that automatically for you can change the formula manually.)

Answer 1

given

Fd = 0.5*rho*u^2*Cd*A

Cd = 0.55

2*pi*r = 12 in

hence, cross section area

A = pi(12 in/2*pi)^2 = 11.45915 in^2 = 0.00739298 m^2

assume m = 6.5 ounce = 0.184272 kg

v = 100 ft/s = 30.48 m/s

a. there3 will be two forces acting on the ball

in -ve x directiuon the horizontal compoennt of drag = Fdx

in -ve y direction initially and then +y direction when the ball is comingd down

b. along x axis

0.5*rho*sqrt(vx^2 + vy^2)vx*Cd*A = m*dvx/dt

along y axis for going up

mdvy/dt = -0.5*rho*sqrt(vx^2 + vy^2)*vy*Cd*A - mg

fgor going down

mdvy/dt = mg - 0.5*rho*sqrt(vx^2 + vy^2)*vy*Cd*A

c. to solve these equations

first take vx^2 = v^2*cos^2(30)

take vy^2 = v^2*sin^2(30)

find dvy/dt =slopey

dvx/dt = slopex

vy(n+1)= vy(n) + (-0.5*rho*sqrt(vx^2 + vy^2)*vy*Cd*A - mg)*dt/m

vx(n+1) = vx(n) + (0.5*rho*sqrt(vx^2 + vy^2)vx*Cd*A)*dt/m

d. in excel

define

m = 0.184272

A = 0.00739298

Cd = 0.55

rho = 1.125

vyo = 30.48*sin(30) = 15.24

vxo = 30.48*cos(30) = 26.39

g = 9.81

dt = 0.001 s

vy(n+1) = vy(n) - (0.5*rho*sqrt(vx(n)^2 + vy(n)^2)&vy(n)*Cd*A - mg)dt/m

vx(n+1) = vx(n) + 0.5*rho*A*Cd*vx(n)*sqrt(vx(n)^2 + vy(n)^2)*dt/m