## College Physics (7th Edition)

Published by Pearson

# Chapter 3 - Motion in Two Dimensions - Learning Path Questions and Exercises - Exercises - Page 102: 88

#### Answer

a). Sketches of views on the people from deck (left) and the people on the boat with respect to the water is shown. b). Relative to the ship, camera's initial velocity is zero. Relative to water, velocity is that of the ship, 2.4m/s due North. c). h=10m, $h=\frac{1}{2}gt^{2}$ $t=\sqrt \frac{2h}{g}=1.4s$ d). The distance traveled cannot be determined as viewed from the passengers on boat in rest. This is because it will resemble a parabola. However, the distance traveled according to the ship's passengers will be h=10m. e). The people on the boat at rest with respect to the water will notice a vertical net displacement of 10m and horizontal net displacement of $x=vt=2.4\times1.4=3.4m$ These displacements are perpendicular, so net displacement = $\sqrt (10^{2}+3.4^{2})\approx10.6m$

#### Work Step by Step

a). Sketches of views on the people from deck (left) and the people on the boat with respect to the water is shown. b). Relative to the ship, camera's initial velocity is zero. Relative to water, velocity is that of the ship, 2.4m/s due North. c). h=10m, $h=\frac{1}{2}gt^{2}$ $t=\sqrt \frac{2h}{g}=1.4s$ d). The distance traveled cannot be determined as viewed from the passengers on boat in rest. This is because it will resemble a parabola. However, the distance traveled according to the ship's passengers will be h=10m. e). The people on the boat at rest with respect to the water will notice a vertical net displacement of 10m and horizontal net displacement of $x=vt=2.4\times1.4=3.4m$ These displacements are perpendicular, so net displacement = $\sqrt (10^{2}+3.4^{2})\approx10.6m$

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