Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 2 - Kinematics in One Dimension - Problems - Page 50: 58

Answer

The stone crosses each other at a distance $2.46$ $m$ from the base of the cliff.

Work Step by Step

Let downward be the positive direction and upward be the negative direction. Say, the stone crosses each other at a distance $x$ from the base of the cliff. The equation of motion : $s=ut+\frac{1}{2}gt^{2}$ For upward motion: $x=9t-\frac{1}{2}gt^{2} \tag{1}$ For downward motion: $(6-x)=9t+\frac{1}{2}gt^{2} \tag{2}$ $x=6-9t-\frac{1}{2}gt^{2} \tag{3}$ $9t-\frac{1}{2}gt^{2}=6-9t-\frac{1}{2}gt^{2}$ $18t=6$ Equation equation $(1)$ and $(3)$, $t=\frac{1}{3}$ $s$ Putting $t=\frac{1}{3}$ $s$ in equation $(1)$, $x=9\times\frac{1}{3}-\frac{1}{2}\times9.8\times\frac{1}{3^{2}}$ $x=3-0.54$ $x=2.46$ $m$
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