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: 48a


t = 1.53 seconds

Work Step by Step

The initial height of the roof (and the stone) is 30m. The stone is thrown downward with a speed of 12.0 m/s, therefore the velocity = -12.0 m/s (we designate upwards as positive). We use the equation: $x = x0 + v0*t + (1/2)at^2 $ where x0 and v0 are initial position and initial velocity, respectively The acceleration in this case is that due to gravity, and that is $a = -9.81 m/s^2$ (negative because it is downwards, which we designate as negative) We are solving for t when x = 0 (when the stone hits the ground), therefore: $0 = 30m + (-12m/s)*t +(1.2)(-9.81m/s^2)t^2 $ $0 = 30 -12t -4.905t^2 $ We now have a quadratic, which we can either graph to find the roots or use the quadradic equation: x = $ \frac{-b + \sqrt(b^2 - 4ac) }{2a}$ and x = $ \frac{-b - \sqrt(b^2 - 4ac) using a = -4.905, b = -12, c = 30 We then get two roots, but we take only the positive one since time can only be positive and we get that t = 1.53 seconds
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