Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 4 - Two-Dimensional Kinematics - Problems and Conceptual Exercises: 20

Answer

$\textbf{(a)}$ The angle is $-66^\circ$ with respect to the horizontal. The speed is $v_1=8.1\text{ m/s}$. $\textbf{(b)}$ The angle is $-76^\circ$ with respect to the horizontal. The speed is $v_2=13.7\text{ m/s}$.

Work Step by Step

For both parts, the horizontal component of the velocity will remain constant - $v_x=3.3\text{ m/s}$. $\textbf{(a)}$ Since the vertical component of the initial velocity is equal to zero, the vertical component after $t_1=0.75\text{ s}$ is $$v_{y_1}=gt_1.$$ Now, the vertical and the horizontal component make a right triangle, and the angle between $\vec{v}$ and the horizontal satisfies: $$\tan\theta_1=\frac{v_{y_1}}{v_x}\Rightarrow\theta_1=\arctan\left(\frac{v_{y_1}}{v_x}\right).$$ Plugging in for $v_{y_1}$ we get $$\theta_1=\arctan\left(\frac{gt_1}{v_x}\right)=\arctan\left(\frac{9.81\text{ m/s}^2\cdot0.75\text{ s}}{3.3\text{ m/s}}\right)=66^\circ.$$ Because the angle is below the horizontal it would be more appropriate to say that the direction is at $-66^\circ$ with respect to the horizontal. As for the magnitude, by Pythagorean theorem, we have $$v_1=\sqrt{v_x^2+v_{y_1}^2}=\sqrt{v_x^2+g^2t_1^2}=\sqrt{3.3^2+9.81^2\cdot0.75^2}\text{ m/s}=8.1\text{ m/s}.$$ $\textbf{(b)}$ The final vertical component of the velocity is (because the initial is equal to zero): $$v_{y_2}=\sqrt{2gh}$$ where $h=9.0\text{ m}$. Plugging this into the expresson for the angle we get $$\theta_2=\arctan\left(\frac{v_{y_2}}{v_x}\right)=\arctan\left(\frac{\sqrt{2gh}}{v_x}\right)$$ which is, finally: $$\theta_2=\arctan\left(\frac{\sqrt{2\cdot9.81\text{ m/s}^2\cdot 9\text{ m}}}{3.3\text{ m/s}}\right)=76^\circ.$$ Because the angle is below the horizontal, it would be more appropriate to say that the direction is at $-76^\circ$ with respect to the horizontal. As for the magnitude, by Pythagorean theorem, we have $$v_2=\sqrt{v_x^2+v_{y_1}^2}=\sqrt{v_x^2+2gh}=\sqrt{3.3^2+2\cdot9.81\cdot9.0}\text{ m/s}=13.7\text{ m/s}.$$
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