Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 4 - Motion in Two and Three Dimensions - Problems: 39d

Answer

Angle is below the horizontal line

Work Step by Step

Lets start by calculating the horizontal component of this object which is constant during motion: $v_{x}=\dfrac {d}{t}=\dfrac {25m}{1.5s}\approx 16.67\dfrac {m}{s}$ Now lets calculate the vertical component of velocity when the object hits the ground. We can use $\theta$ and $v_{x}$ to calculate vertical component: $\dfrac {v_{y}}{v_{x}}=\tan \theta \Rightarrow v_{y}=v_{x}\tan \theta \approx 16.67\dfrac {m}{s}\times \tan -60^{0}\approx —28.87\dfrac {m}{s}$ So we can find initial vertical velocity of object as: $v_{y}=v_{0y}-gt\Rightarrow -28.87=v_{0y}-1.5\times9.8\dfrac {m}{s^{2}}\Rightarrow \overrightarrow {v_{0}}_{y}\approx -14.17\dfrac {m}{s}$ The (-) sign shows that the object moves in the (-) y direction. We can calculate the initial angle $\theta_{0} $ easily using the equation: $\dfrac {v_{0y}}{v_{0x}}=\tan \theta _{0}\Rightarrow \theta _{0}=\arctan \dfrac {v_{0y}}{v_{0x}}\approx -40.37^{0}$ The (-) sign means it is below the horizontal line.
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