## Essential University Physics: Volume 1 (4th Edition)

First, we set equation 13.4 in the textbook equal to zero to obtain: $(\frac{g}{2(v_0^2)})cos^2{\theta}x^2-tan\theta x + y = 0$ Plugging in the known values, it follows: $x=3.1,8.7$ Since x=3.1 is simply where it crosses the corner, we know we want the second value of x. Thus, we find that the distance it lands from the cliff is: $=8.7-3.1=5.6 \ m$