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PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, 
you are using it without permission. 
213 
 
 
PROBLEM 11.144 
A basketball is bounced on the ground at Point A and rebounds with a velocity Av of 
magnitude 2.5 m/s as shown. Determine the radius of curvature of the trajectory 
described by the ball (a) at Point A, (b) at the highest point of the trajectory. 
 
SOLUTION 
 

 
(a) We have 
2
( ) A
A n
A
v
a
ρ
= 
 or 
2
2
(2.5 m/s)
(9.81 m/s ) sin 15
Aρ =
°
 
 or 2.46 mAρ =  
(b) We have 
2
( ) B
B n
B
v
a
ρ
= 
 where Point B is the highest point of the trajectory, so that 
 ( ) sin 15B A x Av v v= = ° 
 Then 
2
2
[(2.5 m/s) sin 15°]
0.0427 m
9.81 m/s
Bρ = = 
 or 42.7 mmBρ = 

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