Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 3 - Kinematics in Two Dimensions; Vectors - Problems: 33

Answer

a. 4.0m/s, 55$^o$ above the horizontal b. 4.6m c. 9.7m/s, 76$^o$ below the horizontal

Work Step by Step

a. Let x= 0 and y = 0 directly below where the diver jumps, and let upward be the positive y direction. To find the initial velocity, find the 2 components of the velocity at that time. Find the initial horizontal velocity. The horizontal distance covered is the constant horizontal velocity multiplied by the time. $$(Vx)(1.3 s) = 3.0 m$$ We find that Vx = 2.308 m/s. Now use equation 2.11b to relate the projectile’s initial vertical velocity vy to the time of 1.3 s and initial and final positions. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. The diver starts at a height of y = 4.0 m, and her final vertical position is 0 m. The initial y-velocity is 3.293 m/s. $v=\sqrt {(2.308m/s)^2+(3.293m/s)^2}$ = 4.0m/s. The initial velocity is 4.0m/s. b. The maximum height is found using the initial y-velocity, and realizing that the vertical velocity at the top is zero. Use Equation 2.11c with the values above. $y_{max}=4.0m-\frac{(3.293m/s)^2}{2(-9.8m/s^2)}$ = 4.55m.The maximum vertical displacement is 4.6 m. c. To find the final velocity, find the 2 components of the velocity at that time. The final horizontal velocity is the same as the initial, Vx = 2.308 m/s. The final vertical velocity is found using equation 2.11a, with a time of 1.3 s. It is -9.447 m/s, directed downward as expected. $v=\sqrt {(2.308m/s)^2+(-9.47m/s)^2}$ = 9.7m/s
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