Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 2 - Motion Along a Straight Line - Problems: 47a



Work Step by Step

The pipe strikes the ground at a speed of 24m/s. The acceleration due to gravity has a magnitude of $9.8m/s^2$. This means that each second, the velocity increases by 9.8m/s. Since the wrench was dropped, the initial velocity was 0m/s. We use the equation: $v =v0 + at$ Where v0 = initial velocity = 0 We get $24m/s = (9.8 m/s^2)t $ and solving for t we get t = 2.45 seconds Now that we have the time it took to reach the ground, we can use the equation $x = x0 +v0*t + (1/2)at^2$ This can be re-written as $\Delta x = v0*t +(1/2)at^2 $ Once again, v0 = 0, $a = 9.81 m/s^2$ and we now know t = 2.45s Using all this information we have $\Delta x = 29.4m$
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