Answer
The cylinder hits the ground a horizontal distance of $~~4.0~m~~$ from the roof's edge.
Work Step by Step
In part (a), we found that the speed when the cylinder leaves the roof is $6.26~m/s$
We can find the vertical component of the velocity:
$v_y = -(6.26~m/s)~sin~30^{\circ}$
$v_y = -3.13~m/s$
We can find the time it takes the cylinder to fall to the ground:
$H = v_{0y}~t+\frac{1}{2}a_yt^2$
$-5.0~m = (-3.13~m/s)~t+\frac{1}{2}(-9.8~m/s^2)t^2$
$4.9t^2+3.13t-5.0 = 0$
We can use the quadratic formula:
$t = \frac{-3.13\pm \sqrt{(-3.13)^2-(4)(4.9)(-5.0)}}{(2)(4.9)}$
$t = \frac{-3.13\pm \sqrt{107.8}}{9.8}$
$t = -1.38~s, 0.74~s$
Since time $t$ is positive, the solution is $t = 0.74~s$
We can find the horizontal distance the cylinder travels in this time:
$x = v_x~t$
$x = (6.26~m/s)(cos~30^{\circ})(0.74~s)$
$x = 4.0~m$
The cylinder hits the ground a horizontal distance of $~~4.0~m~~$ from the roof's edge.
