Speed of the cannon is 95 m/s
now we will find its two components
speed in horizontal direction is
[tex]v_x = 95 cos15[/tex]
[tex]v_x = 91.76 m/s[/tex]
speed in vertical direction is
[tex]v_y = 95 sin(-15)[/tex]
[tex]v_y = -24.6 m/s[/tex]
now the time taken by the cannon to reach the water surface will be find out by using kinematics
[tex]\Delta y = v_y * t + \frac{1}{2} at^2[/tex]
[tex]-3 = -24.6 * t - \frac{1}{2}9.8 * t^2[/tex]
[tex]4.9t^2 + 24.6 t - 3 = 0[/tex]
By solving above quadratic equation
[tex]t = 0.12 s[/tex]
now in the same time the horizontal distance moved by the cannon
[tex]d_x = v_x * t[/tex]
[tex]d_x = 91.76 * 0.12[/tex]
[tex]d_x = 10.93 m[/tex]
Since the distance of crewmen is 20 m so it will definitely safe as cannon will hit at distance 10.93 m
