A baseball pitcher throws a curveball with an initial velocity of 65 mi/h and a spin of 6500 r/min about a
vertical axis. A baseball weighs 0.32 lbf and has a diameter of 2.9 in. Using the data of Fig. P7.108 for
turbulent flow, estimate how far such a curveball will have deviated from its straight-line path when it
reaches home plate 60.5 ft away.
a. approximate solution
lb.s/ft2
ax=-22 ft/s2 and x=60.5 ft => x=1/2axt2+v0t => t=0.69 sec => v= axt+v0=79.8 ft/s
V2
R  ave 
ay
(
95  79.8 2
)
 x 
 60.5 
o
2
 898.9 ft;   sin 1    sin 1 
  3.86
8.5
R
898
.
9
 


y  R1  cos    898.91  cos 3.86  2.03 ft
b. exact solution
Ft  mat  CD
D 2
1
dV (t )
dV
ACD
1
ACD
AV 2 (t )  m
 2 
dt    
t  Const
2
dt
V
2m
V
2m


4

  0.00238 slug / ft 3 

W 0.32
1
1

m

 0.0099    0.00243t  0.0105
or : V 
g 32.2
0.00243t  0.0105
 V

CD  0.44

mi
ft 
V (t  0)  65
 95
h
s 
1
V 2 (t )
2m
[1  (dy / dx) 2 ]3 / 2
2m
Fn  man  C L AV 2 (t )  m
r


2
2
2
r
ACL
d y / dx
ACL
A
 0.046 ft 2
dy / dx   
[1   2 ]3 / 2
2m
d
ACL

ACL



dx 

x  Const
2 3/ 2
2
d / dx
ACL
[1   ]
2m
2m
1 
  ( x  0)  dy / dx( x  0)  0 

1  2

ACL
2m
dy / dx 2
x
2
1  dy / dx 
 ACL 

x
 2m 
 ACL  2

 x
 2m 
  ACL  2 
 ACL  2
2
 1  
x dy / dx   
dx

 x  dy 
  2m  
2
2
m






1   ACL  x 2 
  2m  


2
 2m 

 y  
 ACL 
2
  ACL  2 2 
1  
x   Const
  2m  


 2m 

y ( x  0)  0  y  

AC
L 

at x=60.5: y=1.722 ft
  ACL  2 2   2m 
1  

x  
  2m    ACL 


2