Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 3 - Exercises and Problems - Page 49: 80


a) $\theta = 60.74^{\circ}$ b) $v_0=\sqrt{\frac{5g}{sin^2(60.74)}}=8.03 \ m/s$

Work Step by Step

We use the equation for range and for the maximum height reached to find: $2.5 = \frac{v_0^2sin^2\theta}{2g}$ $2.8 = \frac{v_0^2sin2\theta}{g}$ Thus, we find: $\frac{v_0^2sin^2\theta}{5g}=\frac{v_0^2sin2\theta}{2.8g}$ $.2sin^2 \theta = .357 sin\theta cos\theta$ $\theta = 60.74^{\circ}$ This means that the value of the initial speed is: $v_0=\sqrt{\frac{5g}{sin^2(60.74)}}=8.03 \ m/s$
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