## Physics: Principles with Applications (7th Edition)

Published by Pearson

# Chapter 3 - Kinematics in Two Dimensions; Vectors - Problems: 37

#### Answer

a. $4.0\times10^1m/s$ b. $24m/s$

#### Work Step by Step

a. Choose x = 0 and y = 0 to be the point of launch with speed v and initial angle $\theta = 0$, and upward to be the positive y direction. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. When the stuntperson clears the 8 cars, the position is (22 m, -1.5 m). Find the time to clear the cars from the horizontal motion at constant velocity. $$t = \frac{22 m}{(v cos 0)}$$ Plug this expression for the time into equation 2.11b and analyze the vertical motion. y-component of initial velocity: $$(v)(sin 0 ^{\circ}) = 0 m/s$$ We know the final and initial heights, the initial vertical velocity, and the acceleration. Solve for the initial v = 40 m/s or $4.0\times10^1m/s$. b. Start this problem the same way as part (a). $$t = \frac{22 m}{(v cos 7.0 ^{\circ})}$$ Now the y-component of initial velocity is: $$(v)(sin 7 ^{\circ})$$ The vertical displacement is the same, and equation 2.11b is once again used. This yields a starting speed of $v = 24 m/s$.

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