## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 3 - Exercises and Problems - Page 49: 83

#### Answer

The proof is below.

#### Work Step by Step

We know that equation 3.14 is given by: $$y = x tan \theta_0 - \frac{g}{ v_o^2cos^2 \theta}x^2$$ From this, we find that dy/dx is: $$\frac{dy}{dx} = tan \theta_0 - \frac{g}{ v_o^2cos^2 \theta}x$$ We know that dy/dx gives the slope, along with $tan \theta$. Thus, we know: $$tan\theta = \frac{dy}{dx} = tan \theta_0 - \frac{g}{ v_o^2cos^2 \theta}x$$ We now consider equations 3.10 and 3.11: $$v_y = v_{y0}-gt$$ $$v_x = v_{x0}$$ We simplify to find: $$v_y = v_{0}sin \theta_0-gt$$ $$v_x = v_{0}cos \theta_0$$ Since $v_y = dy$ and $v_x=dx$, we find: $$\frac{v_y}{v_x}=\frac{dy}{dx}=tan \theta = \frac{ v_{0}sin \theta_0-gt}{v_{0}cos \theta_0}$$ We finally simplify to: $$tan\theta = \frac{dy}{dx} = tan \theta_0 - \frac{g}{ v_o^2cos^2 \theta}x$$ Thus, we conclude that $dy/dx$ is, indeed, the velocity.

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