Answer:
The safety net should be 13.65 m away from the canon.
Explanation:
The location for the safety net should be exactly at the position where the man will hit the ground.
This is a projectile motion.
We should separate the motion into x- and y-components, and investigate separately.
[tex]v_0 = 11.6~m/s\\v_0_x = v_0\cos(42^\circ) = 8.62~m/s\\v_0_y = v_0\sin(42^\circ) = 7.76~m/s[/tex]
On the x-direction, there is no acceleration, so the kinematics relations are
[tex]x = v_0_x t = 8.62t[/tex]
On the y-direction, there is gravitational acceleration, -9.8. We will look for the time that takes for the man to hit the ground. The kinematics relations are
[tex]y - y_0 = v_0_yt -\frac{1}{2}at^2\\0 - 0 = 7.76t - \frac{1}{2}9.8t^2\\0 = 7.76t - 4.9t^2\\t = 0 ~and ~ t = 1.58s[/tex]
Of course, we will not choose t = 0, since we know that t = 0 is the initial time.
We will plug this into the x-direction equation.
[tex]x = 8.62t = 8.62*1.58 = 13.65~m[/tex]
