Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems - Page 115: 82

Answer

$3.76 \;\rm m$

Work Step by Step

We have here two stages of motion, the first one moving up the ramp and the second one moving in the air until he reaches the ground. To find the distance from the ramp, we need to find the velocity just before he leaves the ramp. As you see in the figure below, we need to find the velocity at the edge of the ramp in the $\acute x $ since the initial velocity is also in this direction. $$v_{1\acute x}^2=v_{0\acute x}^2+2a_{\acute x}d$$ whereas $d$ is the distance traveled up the ramp. As we see below, $a_{\acute x}=-g\sin30^\circ$ the negative sign is due to the direction and since it opposes the direction of the man motion. And $d$ is given by $$\sin 30^\circ=\dfrac{1}{d}$$ Hence, $d=2\;\rm m$ Now we need to plug all the known into the first formula above; $$v_{1\acute x} =\sqrt{7^2-2\cdot 9.8\sin30^\circ \cdot 2 }$$ $$v_{1\acute x} =\color{blue}{v_{02}=5.4\;\rm m/s}\tag1$$ Now, in the second stage of motion, we chose the edge of the ramp to be our initial point. $$(x_0,y_0)=(0,1)$$ Now we need to find the time of this trip from the edge of the ramp to the ground. $$y=y_0+v_{0y}t-\frac{1}{2}gt^2$$ $$0=1+v_{02}\sin30^\circ t-4.9t^2$$ $$0=1+5.4\sin30^\circ t-4.9t^2$$ Thus, $t=-0.2536$ s (dismissed) or $t=\bf 0.8046$ s Now we can find the distance from the end of the ramp at which the skateboarder touches down. $$x_f=v_{02}\cos30^\circ t=5.4\cos30^\circ \cdot 0.8046$$ $$x_f=\color{magenta}{\bf 3.76}\;\rm m$$
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