Answer:
4.49m
Explanation:
The vertical and horizontal component of the tile velocity when it exit the roof are
[tex]v_v = vsin(\alpha) = 6sin(30^0) = 3 m/s[/tex]
[tex]v_h = vcos(\alpha) = 6cos(30^0) = 3.89 m/s[/tex]
Let g = 9.8 m/s2. We can use the following equation of motion for the vertical distance landing to calculate the time t it takes to land
[tex]s_v = v_vt + gt^2/2[/tex]
[tex]10 = 3t + 4.9t^2[/tex]
[tex]4.9t^2 + 3t - 10 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{-3\pm \sqrt{(3)^2 - 4*(4.9)*(-10)}}{2*(4.9)}[/tex]
[tex]t= \frac{-3\pm14.32}{9.8}[/tex]
t = 1.15 or t = -1.77
Since t can only be positive we will pick t = 1.15
This is also the time it takes for the tile to travel horizontally, we can use this to calculate how far from the exit point on the roof that it lands
[tex]s_h = v_ht = 3.89*1.15 = 4.49 m[/tex]
