University Physics with Modern Physics (14th Edition)

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

Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 61: 2.44

Answer

a) The position and velocity of the sandbag from the ground at $t=0.250s$ is $40.9m$ and $2.55\frac{m}{s}$. The position and velocity of the sandbag from the ground at $t=1.00s$ is $40.1m$ and $-4.80\frac{m}{s}$. b) The sandbag strikes the ground $3.41s$ after its release. c) The magnitude of velocity with which the sandbag strikes the ground is $28.4\frac{m}{s}$ d) The maximum height above the ground which the sandbag reaches is $41.3m$. e) See the graph below.

Work Step by Step

We take the ground to be the origin of our system with (+)ve y-direction upwards and let us start measuring $(t=0)$ when we drop the sandbag at a height of $40.0m$ above the ground . Since the motion is uniform, we can use the kinematics equations. a) Let us calculate the position and velocity of the sandbag at $t=0.250s$ and $t=1.00s$. i) At $t=0.250s$ $y-y_{0}=v_{0y}t+\frac{1}{2}a_{y}t^2$ $y=40.0+5.00\times0.250-\frac{1}{2}\times9.80\times0.250^2=40.9m$ and $v_{y}=v_{0y}+a_{y}t=5.00-9.80\times0.250=2.55\frac{m}{s}$ The position and velocity of the sandbag from the ground at $t=0.250s$ is $40.9m$ and $2.55\frac{m}{s}$. ii) At $t=1.00s$ $y-y_{0}=v_{0y}t+\frac{1}{2}a_{y}t^2$ $y=40.0+5.00\times1.00-\frac{1}{2}\times9.80\times1.00^2=40.1m$ and $v_{y}=v_{0y}+a_{y}t=5.00-9.80\times1.00=-4.80\frac{m}{s}$ The position and velocity of the sandbag from the ground at $t=1.00s$ is $40.1m$ and $-4.80\frac{m}{s}$ b) Let us now calculate the time taken by the sandbag to strike the ground after its release. $y-y_{0}=v_{0y}t+\frac{1}{2}a_{y}t^2$ $=>\frac{1}{2}\times9.80t^2-5.00t-40.0=0$ Using quadratic formula, we get $t=\frac{-(-5.00)+-\sqrt ((-5.00)^2-4\times\frac{9.80}{2}\times(-40.0))}{2\times\frac{9.80}{2}}=3.413s\approx3.41s$ [We take the (+)ve time] The sandbag strikes the ground $3.41s$ after its release. c) Let us calculate the velocity with which the sandbag strikes the ground. $v_{y}=v_{0y}+a_{y}t=5.00-9.80\times3.413=-28.4\frac{m}{s}$ The magnitude of velocity with which the sandbag strikes the ground is $28.4\frac{m}{s}$ d) Let us calculate the maximum heigth above the ground which the sandbag reaches. $2a_{y}(y-y_{0})=v_{y}^2-v_{0y}^2$ , $v_{y}=0$ at peak height $y=y_{0}-\frac{v_{0y}^2}{2a_{y}}=40.0-\frac{5.00^2}{-2\times9.80}=41.3m$ The maximum height above the ground which the sandbag reaches is $41.3m$ . e)See the graph of $a_{y}-t, v_{y}-t$ and $y-t$.
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