Consider the problem of a hard-hit baseball. The air-friction drag on a baseball is approximately given by the following formula

and subsequent differential equations :

**F:=-((C_d)*rho*Pi*(r)^2*v*v)/2;**

**m:=0.145;**

**v0:=65;**

**g:=9.81;**

**v_x:=diff(x(t),t);**

**v_y:=diff(y(t),t);**

**d2v_x:=-((C_d)*rho*Pi*(r^2)*(v_x)*sqrt((v_x)^2 +(v_y)^2))/(2*m);**

**d2v_y:=-((C_d)*rho*Pi*(r^2)*(v_y)*sqrt((v_x)^2 +(v_y)^2))/(2*m)-g;**

where

C[d] is the drag coefficient (about 0.35 for a baseball)

rho[air] is the density of air (about 1.2 kg/

3m

)

r is the radius of the ball (about 0.037 m)

v is the vector velocity of the ball

Then if given that :

Power hitters in baseball say they would much rather play in Coors Field in Denver than in sea-level stadiums because it is so much easier to hit home runs. The air pressure in Denver is about 10% lower than it is at sea level. The field dimensions at Coors Field are:

Left Field - 347 feet (106 m)

Left-Center - 390 feet (119 m)

Center Field - 415 feet (126 m)

Right-Center - 375 feet (114 m)

Right Field - 350 feet (107 m)

** 1. Overlay two plots: one at sea level and one in Denver to show why power hitters prefer Coors field.**

**2. Find the initial magnitude of velocity, v0**

**needed to hit a home run to Right-Center, where v_x(0)=v0/sqrt(2) and v_y(0)=v0/sqrt(2)**

I don't quite understand how to use the field dimensions for both 1 and 2 and am pretty clueless as to how to approach this question using the ordinary differential equations mentioned above.