Answer
5.6 meters
Work Step by Step
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:
$=8.7-3.1=5.6 \ m$
