University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 3 - Motion in Two or Three Dimensions - Problems - Exercises - Page 97: 3.61

Answer

(a) Speed of the dog is, $v_{dog}=4.25 m/s$ (b) Horizontal distance travelled by the ball, $R \approx10.58 m$

Work Step by Step

(a) For dog to catch the ball, he must has same speed as horizontal speed of the ball, So, speed of the dog is, $v_{dog}=v_{x, ball}=8.5\times cos60^0=4.25 m/s$ (b) Using $2^{nd}$ equation of motion in vertical direction (taken upward direction as positive), $s_y=u_y t+\frac{1}{2}a_yt^2\Rightarrow -12=8.5sin60^0 t+\frac{1}{2}(-9.8)t^2$ $\Rightarrow 4.9t^2-7.36t-12=0$ Positive solution of this quadratic equation is, $T\approx 2.49 sec$ So, range i.e. horizontal distance travelled by the ball will be, $R=v_x T= = 4.25\times2.49 \approx10.58 m$
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