Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 3 - Kinematics in Two Dimensions - Problems - Page 74: 34

Answer

$(a)\space 85\space m$. $(b)\space 610\space m$

Work Step by Step

Let's find the flight time of the ball on earth. Here we use the equation $S=ut+\frac{1}{2}at^{2}$ in the vertical direction. $\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation. $0=45\space m/s\space \times sin29^{\circ}t+\frac{1}{2}(-9.8\space m/s^{2})t^{2}$ $t=\frac{2\times(21.82\space m/s)}{9.8\space m/s^{2}}=4.45\space s$ The flight time on the distant planet $=3.5\times4.45\space s=15.6\space s$ We know that this is twice the amount of time required for the ball to reach its maximum height, which is 7.80 s (a) Let's apply equation $S=\frac{(u+V)}{2}t$ in the vertical direction to the ball on the distant planet to find its maximum height. $\uparrow S=\frac{(u+V)}{2}t$ ; Let's plug known values into this equation. $H=\frac{45\space m/s\times sin29^{\circ}}{2}\times 7.8\space s=85\space m$ Maximum height = 85 m (b) Let's apply equation $S=ut$ in the horizontal direction to find the range of the ball on the distant planet. $\rightarrow S=ut$ $R=45\space m/s \times cos29^{\circ}\times 15.6\space s=610\space m$ Range = 610 m
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