479 Horizontal Motion: The horizontal component of velocity ... - Wuala

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 487
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*R1–12. The skier starts fom rest at A and travels down
the ramp. If friction and air resistance can be neglected,
determine his speed v B when he reaches B. Also, compute
the distance s to where he strikes the ground at C, if he
makes the jump traveling horizontally at B. Neglect the
skier’s size. He has a mass of 70 kg.
50 m
A
B
v B
4 m
Potential Energy: The datum is set at the lowest point B. When the skier is at point
A, he is (50 - 4) = 46 m above the datum. His gravitational potential energy at this
position is 70(9.81) (46) = 31588.2 J.
s
C
30
Conservation of Energy: Applying Eq. 14–21, we have
T A + V A = T B + V B
0 + 31588.2 = 1 2 (70) y2 B
y B = 30.04 m>s = 30.0 m>s
Ans.
Kinematics: By considering the vertical motion of the skier, we have
( +T) s y = (s 0 ) y + (y 0 ) y t + 1 2 (a c) y t 2
4 + s sin 30° = 0 + 0 + 1 (9.81) t2
2
[1]
By considering the horizontal motion of the skier, we have
A ; + B
s x = (s 0 ) x + y x t
s cos 30° = 0 + 30.04 t
[2]
Solving Eqs. [1] and [2] yields
s = 130 m
Ans.
t = 3.753 s
487