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2-12c with x replaced by y. 0 m ) (b) For the acceleration to be smaller, in the above equation we see that the displacement should be larger. This means that the net should be “loosened” . ⎛ 1m s ⎞ 73. 8 m s . 6 km h ⎠ location at which the deceleration begins. We have v = 0 m s and a = − 30 g = − 294 m s 2 . Find the displacement from Eq. 2-12c. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 59 × 10 4 m H If the larger answer is used in t1 = T − , a negative time of fall results, and so the physically vs 2 correct answer is H = 52 m . 66. (a) Choose up to be the positive direction. Let the throwing height of both objects be the y = 0 location, and so y0 = 0 for both objects.

Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 3m 74. Choose downward to be the positive direction, and y 0 = 0 to be at the start of the pelican’s dive. 0 m. Find the total time of the pelican’s dive from Eq. 2-12b, with x replaced by y. 81 s . 61 s into the dive. Find the location of the pelican at that time from Eq.